Ionic Equillibrium

IONIC EQUILLIBRIA

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ACIDS AND BASES

The Nature of Acids and Bases Arrhenius concept:

Acid a substance that produces hydrogen ion in aqueous solution.

Base a substance that produces hydroxide ion in aqueous solution. Examples of Acids:

1. HCl(aq) H+(aq) + Cl-(aq); (a strong acid)

2. HNO3(aq) H+(aq) + NO3

-(aq); (a strong acid)

3. CH3COOH(aq) ⇋ H+(aq) + CH3COO-

(aq); (a weak acid) Examples of bases:

1. NaOH(aq) Na+(aq) + OH-

(aq); (a strong base)

2. Ba(OH)2(aq) Ba2+(aq) + 2OH-

(aq); (a moderately strong base)

8. NH3(aq) + H2O ⇋ NH4+

(aq) + OH-(aq); (a weak base)

* All hydroxides and oxides of metal are bases. (Oxides of metals in veryhigh O.N. is acidic. Ex CrO3 , Mn2O7 . The Brønsted-Lowry concept:

Acid a substance that acts as a proton donor in chemical reaction;

Base a substance that acts as a proton acceptor in chemical reaction; * Reactions that involve the transfer of protons (H+) are acid-base reactions. The Bronsted-Lowry acid-base reaction can be represented as follows:

HA + B ⇋ BH+ + A- acid1 base2 conjugate conjugate acid2 base1

For examples:

1. HCl + H2O H3O+

(aq) + Cl-(aq) acid1 base2 conjugate conjugate acid2 base1

2. HC2H3O2 + H2O ⇋ H3O+

(aq) + C2H3O2-(aq);

acid1 base2 conjugate conjugate acid2 base1

3. NH3 + H2O ⇋ NH4+

(aq) + OH-(aq);

base1 acid2 conjugate conjugate acid1 base2 In each reaction, a conjugate base is what remains of the acid molecule after it loses a proton (H+), and a conjugate acid is what becomes of the base after it gains a proton. acid1 - conjugate base1 and acid2 - conjugate base2 are called conjugate acid-base pairs; they are a pair of species that are related to each other by the loss or gain of a single

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MASTERING PHYSICAL CHEMISTRY By-- S.K.SINHA

H.O-2 N 12 Rangbari Yojana , Kota B.O-1D 10 Talwandi main Road, Kota Ph-0744-2407906 Mo-93149-05055.

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proton (H+). Thus, H2O and H3O+ are conjugate acid-base pair; H3O

+ is called hydronium ion. A BrØnsted-Lowry acid-base reaction involves a competition between two bases for a proton, in which the stronger base ends up being the most protonated at equilibrium. In the reaction: HCl + H2O H3O

+(aq) + Cl-(aq),

H2O is a stronger base than Cl-; at equilibrium, a solution of hydrochloric acid contains mostly H3O

+ and Cl- ions.

In reaction: HC2H3O2 + H2O ⇋ H3O+

(aq) + C2H3O2-(aq);

C2H3O2- is the stronger base; at equilibrium, a solution of acetic acid contains mostly

HC2H3O2 and a small amount of H3O+ and C2H3O2

- ions.

In the reaction: NH3 + H2O ⇋ NH4+

(aq) + OH-(aq),

water is an acid. The competition for protons is between NH3 and OH-, in which OH- is the strong base; the above equilibrium shifts to the left. That is, an aqueous ammonia solution contains mostly NH3 molecule and a small amount of NH4OH, NH4

+, and OH-. Exercises:

1. Write the conjugate base for each of the following acids:

(a) H2PO4-; (b) H2C2O4; (c) [Al(H2O)6]

3+; (d) NH3;

2. Write the conjugate acid for each of the following bases:

(a) NH3 (b) [Al(H2O)2(OH-)4]- (c) SO3

2- (d) (CH3)3N:

The fects of Structure of Acid-Base Properties

The Relative Strength of Acids

The strength of acids is determined by a combination of various factors, such as the strength and polarity of XH bond in the molecule, and the hydration energy of the ionic species in aqueous solution. For binary acids such as HF, HCl, HBr, and HI, HX bond strength decreases down the group. The weaker the bond, the easier the molecule ionizes in aqueous solution. Hence, the stronger will be the acid. Thus, for this group of acids, their strength increases down the group: HF < HCl < HBr < HI; H2O < H2S < H2Se < H2Te Among the hydrohalic acids, HF is the only weak acid; the others are strong acids. The relative strength of HCl, HBr, and HI cannot be differentiated in aqueous solution, because each of them dissociates almost completely. Less polar solvents are used to determine their relative strength. For example, HCl, HBr and HI ionize only partially in acetone or methanol, which have a weaker ionizing strength than water. The ionization of HCl in acetone can be represented by the following equilibrium:

(CH3)2CO(l) + HCl(acetone) ⇋ (CH3)2COH+(acetone) + Cl-(acetone)

The degree of ionization in acetone increases in the order of HCl < HBr < HI. Based on this observation, it was determined that the acidity of hydrogen halides increases down the



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group: HF < HCl < HBr < HI. A similar trend of relative acidity is also observed for the hydrides of Group VIA elements, such that, H2O < H2S < H2Se < H2Te. For the same period hydrides, the relative acidity increases from left to right, such that:

CH4 < NH3 < H2O < HF; PH3 < H2S << HCl Water is a stronger acid than ammonia, and in acid-base reaction, H2O acts as a BrØnsted-Lowry acid, which donates a proton to NH3:

H2O(l) + NH3(aq) ⇋ NH4+

(aq) + OH-(aq)

In reaction with HF, water acts as a BrØnsted-Lowry base, which accept a proton:

HF(aq) + H2O(l) ⇋ H3O+

(aq) + F-(aq)

The Relative Strength of Oxo-acids Oxo-acids are acids that contain one or more OH groups covalently bonded to a central atom, which can be a metal or a nonmetal. The OH group ionizes completely or partially in aqueous solution producing hydrogen ions. Some common examples of oxo-acids are H2CO3, HNO3, H3PO4, H2SO4, HClO4, and HC2H3O2: O O O HOSOH HOPOH OClOH O O O H

For these type of acids, their relative strengths depend on the electronegativity of the central atoms. The more electronegative the central atom, the more polarized the OH bond and the more readily it ionizes in aqueous solution to release the H+ ion. For example, N, S, and Cl are more electronegative than P, and HNO3, H2SO4, and HClO4 are stronger acids compared to H3PO4. The relative acid strengths are: HNO3 > H2SO4 > H3PO4. For oxo-acids with central atoms from the same group, their relative strength decreases from top to bottom as the electronegativity of the central atom decreases: HOCl > HOBr > HOI; HOClO > HOBrO > HOIO; HOClO2 > HOBrO2 > HOIO2 For oxo-acids containing identical central atom, acidity increases as more oxygen atoms are bonded to it. For example, acidity increases in the following order: HOCl < HOClO < HOClO2 < HOClO3; H2SO4 > H2SO3; HNO3 > HNO2

Oxygen is a very electronegative atom; when more oxygen atoms are bonded to the central atom, the OH group in the molecule becomes more polarized (due to inductive effect), and the more readily it ionizes to release H+ ion.

Acetic acid is an organic acid, which contains the acidic carboxyl group, -COOH. When acetic acid (CH3COOH) ionizes, only the OH bond of the carboxyl group breaks, but not the CH bonds. However, if one or more of the hydrogen atoms in the methyl group (-CH3) is substituted with a more electronegative atom, the OH of -COOH group becomes more polarized and ionizes more readily, and increasing the acidity. The following Ka values illustrate the effect on the acidity of COOH when one or more of the methyl hydrogen is substituted with more electronegative atoms:

CH3COOH(aq) ⇋ CH3COO-(aq) + H+

(aq); Ka = 1.8 x 10-5



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ClCH2COOH(aq) ⇋ ClCH2COO-(aq) + H+

(aq); Ka = 1.4 x 10-3 chloroacetic acid

FCH2COOH(aq) ⇋ FCH2COO-(aq) + H+

(aq); Ka = 2.6 x 10-3 fluoroacetic acid

CCl3COOH(aq) ⇋ CCl3COO-(aq) + H+

(aq); Ka = 3.0 x 10-1 trichloroacetic acid Exercise: 1. Rank the following acids in order of increasing strength:

(a) CH3COOH, CH3CH2COOH, HCOOH;

(b) CH3COOH, ClCH2COOH, CHCl2COOH, CCl3COOH, CF3COOH 2. Rank the following bases in order of increasing strength:

(a) NH3, CH3NH2, (CH3)2NH2, (CH3)3N, (CH3CH2)3N ________________________________________________________________________

Acid-Base Properties of Oxides

Metal oxides are either basic or have amphoteric properties. The oxides the Group IA metals are strongly basic because they are generally very soluble in water. Other metal oxides are less soluble, but they react with strong acids:

Na2O(s) + H2O(l) 2NaOH(aq);

BaO(s) + H2O(l) Ba(OH)2(aq)

MgO(s) + 2HCl(aq) MgCl2(aq) + H2O(l) The oxide ion, O2-, has a very strong affinity for protons and reacts with water to produce hydroxide ions: O2-

(aq) + H2O(l) 2OH-(aq)

Some metal oxides are amphoteric; for example, Al2O3, Cr2O3, PbO, SnO, and ZnO. They react with both strong acids and strong bases:

Al2O3(s) + 6HCl(aq) 2AlCl3(aq) + 3H2O(l);

Al2O3(s) + 2NaOH(aq) + 3H2O(l) 2Na[Al(OH)4](aq); Oxides of nonmetals are acidic. They form acidic solution when dissolved in water.

CO2(g) + H2O(l) ⇋ H2CO3(aq) ⇋ H+(aq) + HCO3

-(aq);

SO2(g) + H2O(l) ⇋ H2SO3(aq) ⇋ H+(aq) + HSO3

-(aq);

P4O10(g) + 6H2O(l) ⇋ 4H3PO4(aq); H3PO4(aq) H+(aq) + H2PO4

-(aq);

Metal Hydroxides and Hydrides Hydroxides of Group IA metals are strongly basic. For example, LiOH, NaOH, and KOH, are strong bases. The basicity of other metal hydroxides is limited by their low solubility. The hydroxides of some metals exhibit amphoteric properties. For example, aluminum hydroxide, Al(OH)3 is an amphoteric hydroxides.

Al(OH)3(s) + OH-(aq) ⇋ Al(OH)4

-(aq);

Al(OH)3(s) + 3H3O+

(aq) ⇋ [Al(H2O)6]3+

(aq)

Cr(OH)3, Zn(OH)2, Sn(OH)2, and Pb(OH)2 are also amphoteric hydroxides.



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The hydrides of reactive metals, such as NaH and CaH2, are strong bases. The hydride ion has a strong affinity for protons and reacts with water to produce hydroxide ions and hydrogen gas: H-

(aq) + H2O(l) H2(g) + OH-(aq)

The Lewis Acids and Bases

According to G.N. Lewis, an acid is the reactant that is capable of sharing a pair of electrons from another reactant to form a covalent bond; a base is the reactant that provides the pair of electrons to be shared to form a covalent bond. Based on the Lewis�s definition, hydrogen ion (H+) is considered a Lewis acid and water and ammonia are the Lewis bases in the following reactions:

H+ + H2O H3O+; H+ + NH3 NH4

+; Lewis Lewis Lewis Lewis acid base acid base Species with incomplete octet (or electron deficient species) may act as Lewis acid and those with lone pair of electrons may act as Lewis bases. For example, in the following reactions, BF3, AlCl3, and FeBr3 are Lewis acids; while NH3, Cl-, and Br- are Lewis bases. BF3 + NH3 F3B:NH3; AlCl3 + Cl- AlCl4

-; FeBr3 + Br- FeBr4-;

In the formation of complex ions, the positive ions act as Lewis acids and the ligands (anions or small molecules) are Lewis bases:

Cu2+(aq) + 4NH3(aq) ⇋ Cu(NH3)4

2+(aq); Lewis acid Lewis base

Al3+(aq) + 6H2O(l) ⇋ [Al(H2O)6]

3+(aq);

Exercise 1. Determine the Lewis acids and Lewis bases in the following reactions:

(a) CO2(g) + H2O(l) ⇋ H2CO3(aq)

(b) SO3(g) + H2O(l) ⇋ H2SO4(aq)

(c) AlCl3 + (CH3)3N: ⇋ (CH3)3N:AlCl3

(d) Zn(OH)2(s) + 2OH-(aq) ⇋ Zn(OH)4

2-(aq);

(e) AuCl3(s) + Cl-(aq) ⇋ AuCl4-(aq) Acids Strength

The strength of an acid is defined by the extent of its dissociation (ionization) in aqueous solution.

HA(aq) + H2O H3O+

(aq) + A-(aq)

Strong acids, such as HClO4, HCl, HNO3, and H2SO4, are those that are considered to completely ionize at 1 M concentration. The above equilibrium would lie far to the right. Acids that are only partially ionized at this concentration, such as HC2H3O2, HF, HNO2, H2SO3, H3PO4, and HClO, are considered weak acids. Their dissociation equilibria would lie far to the left. The equilibrium constant, Ka, for the acid ionization is given by the following expression:

[H3O+][A-] [H+][A-]

Ka = = [HA] [HA]



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For strong acids, Ka would have a very large value; for weak acids, Ka < 10-1. The following are the Ka values of some weak acids:

1. HF(aq) ⇋ H+(aq) + F-

(aq); Ka = [H+][F-] = 7.2 x 10-4

[HF]

2. HNO2(aq) ⇋ H+(aq) + NO2

-(aq); Ka = [H+][NO2

-] = 4.0 x 10-4

[HNO2]

3. HC2H3O2(aq) ⇋ H+(aq) + C2H3O2

-(aq); Ka = [H+][ C2H3O2

-] = 1.8 x 10-5

[HC2H3O2]

4. HOCl(aq) ⇋ H+(aq) + ClO-

(aq); Ka = [H+][ClO-] = 3.5 x 10-8

[HOCl]

5. HCN(aq) ⇋ H+(aq) + CN-

(aq); Ka = [H+][CN-] = 6.2 x 10-10

[HCN] The value of Ka is a measure of the extent of acid dissociation; hence, the strength of the acid. A stronger acids have larger Ka values. For strong acids such as HClO4, HCl, H2SO4, and HNO3, their Ka�s are very large (not given in any tables of Ka values). The Conjugate Acid-Base Relationships

Strong acids have weak conjugate bases, and weak acids have strong conjugate bases; the weaker the acid, the stronger will be its conjugate base. Likewise, weak bases have strong conjugate acids; the weaker the base, the stronger will be its conjugate acid. Examples:

HCl is a strong acid and Cl- is a very weak base;

HF is a weak acid, and F- is a stronger conjugate base than Cl-.

According to BrØnsted-Lowry, acid-base reactions favor the direction from strong acid-strong base combinations to weak acid-weak base combinations:

The following acid-base reactions go completely in the forward direction:

1. HClO4(aq) + H2O(l) H3O+

(aq) + ClO4-(aq); (HClO4 is a very strong acid)

2. HCl(aq) + NH3(aq) NH4+

(aq) + Cl-(aq); (HCl is a strong acid)

3. HSO4-(aq) + CN-

(aq) HCN(aq) + SO42-

(aq); (HSO4- is a stronger than HCN)

Many acid-base reactions reach a state of equilibrium. The following acid-base reactions do not go to completion; their equilibrium positions may shift to the right or to the left, depending on the relative strength of the acid:

1. H2PO4-(aq) + C2H3O2

-(aq) ⇋ HC2H3O2(aq) + HPO4

2-(aq);

Equilibrium shifts left; HC2H3O2 is the stronger acid and HPO42- is the stronger base.

2. HNO2(aq) + C2H3O2-(aq) ⇋ HC2H3O2(aq) + NO2

-(aq);

Equilibrium shifts right; HNO2 is the stronger acid, C2H3O2- is the stronger base.

Water as an Acid and a Base

Water is amphoteric because it can behave as an acid and a base. The following equilibrium occurs in pure water:

H2O + H2O ⇋ H3O+

(aq) + OH-(aq);

Acid1 base1 conjugate conjugate acid2 base1



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The above equilibrium can be simplified as follows, which is also called the auto-ionization of water:

H2O ⇋ H+(aq) + OH-

(aq); Kw = [H+][OH-] = 1.0 x 10-14 at 25 oC.

Kw is called the ion product constant for water. An aqueous solution contains both H+ (in the form of H3O

+) and OH-, such that the product of their concentrations is constant equal to 1.0 x 10-14 at 25 oC. In pure water (or in a neutral solution,

[H+] = [OH-] = 1.0 x 10-7 M. If [H+] increases (> 1.0 x 10-7 M), [OH-] will decreases (< 1.0 x 10-7 M), and vice versa.

Thus, a solution with: [H+] > 1.0 x 10-7 M, or [H+] > [OH-], the solution is acidic;

[H+] < 1.0 x 10-7 M, or [H+] < [OH-], the solution is basic;

[H+] = [OH-] = 1.0 x 10-7 M, the solution is neutral; Exercise-2: 1. What is [OH-] if [H+] = 0.0050 M? Is the solution acidic, basic or neutral? 2. What is the [H+] if [OH-] = 6.0 x 10-4 M? Is the solution acidic, basic or neutral? ________________________________________________________________________

The pH Scale pH is a scale that measures the acidity of an aqueous solution where [H+] is very small.

pH = -log[H+], If [H+] = 1.0 x 10-7 M, pH = -log(1.0 x 10-7) = -(-7.00) = 7.00 We can also express basicity using the log scale for [OH-], such that pOH = -log[OH-]

That is, if a solution has [OH-] = 1.0 x 10-7 M, pOH = -log(1.0 x 10-7) = -(7.00) = 7.00

Since, at 25oC, Kw = [H+][OH-] = 1.0 x 10-14

pKw = -log Kw = -log[H+] + (-log[OH-]) = -log(1.0 x 10-14) = -(-14.00) = 14.00

pKw = pH + pOH = 14.00; and pOH = 14.00 � pH Thus, in aqueous solutions, if pH > 7.00 pOH < 7.00, and vice versa;

In neutral solution, [H+] = [OH-] = 1.0 x 10-7 M, and pH = pOH = 7.00;

[H+] > 1.0 x 10-7 M, pH < 7.00, the solution is acidic.

[H+] < 1.0 x 10-7 M, pH > 7.00, the solution is basic. _________________________________________________________________________.

Calculating the pH of Strong Acid and Strong Base Solutions Strong acids are assumed to ionize completely in aqueous solution. For monoprotic

acids (that is, those acids with a single ionizable hydrogen atom), the concentration of hydrogen ion in solution is the same as the molar concentration of the acid. That is [H+] = [HA] For example, in 0.10 M HCl(aq), [H+] = 0.10 M, and pH = -log(0.10) = 1.00. A strong base such as NaOH, contains OH- which concentration is equal to that of dissolved NaOH. That is, a solution of 0.10 M NaOH(aq) contains [OH-] = 0.10 M pOH = -log[OH-] = -log(0.10) = 1.00; pH = 14.00 � 1.00 = 13.00



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A strong base such as Ba(OH)2 produces twice the concentration of OH- as the concentration of Ba(OH)2 in solution. Ba(OH)2 dissociates as follows: Ba(OH)2(aq) Ba2+

(aq) + 2OH-(aq);

[OH-] = 2 x [Ba(OH)2] In a solution of 0.010 M Ba(OH)2, [OH-] = 0.020 M, pOH = 1.70, and pH = 12.30 Exercise-3: 1. Calculate the pH of the following solutions:

(a) 0.0050 M HCl; (b) 0.00060 M NaOH. 2. What is [H+] and [OH-], respectively, in a solution where,

(a) pH = 4.50; (b) pOH = 5.40 3. The pH of an HCl solution is found to be 3.00. To what final volume must a 100.-mL

sample of this acid be diluted so that the pH of the solution becomes 3.50? (316 mL) 4. What is the pH of a saturated aqueous solution of Ca(OH)2 in which 0.165 g of Ca(OH)2

is dissolved in 100. mL solution at 25 oC? (Answer: pH = 11.65) _________________________________________________________________________.

Calculating the pH of Weak Acid Solutions Weak acids do not ionize completely. At equilibrium, [H+] is much less than the concentration of the acid. The concentration of H+ in a weak acid solution depends on the initial acid concentration and the Ka of the acid. To determine [H+] of a weak acid we can set up the �ICE� table as follows: Consider a solution of 0.10 M acetic acid and its ionization products:

Concentration: HC3H3O2(aq) ⇋ H+(aq) + C2H3O2

-(aq)

Initial [ ], M: 0.10 0.00 0.00

Change, [ ], M: -x +x +x

Equilibrium [ ], M : (0.10 - x) x x The acid ionization constant, Ka, is given by the expression:

Ka = [H+][C2H3O2-] = x2 = 1.8 x 10-5

[HC2H3O2] (0.10 � x) Since Ka << 0.10, we can use the approximation that x << 0.10, and (0.10 � x) ~ 0.10

Then, Ka = x2 ~ x2/0.10 = 1.8 x 10-5 ; (0.10 � x)

x2 = (0.10)(1.8 x 10-5); x = (1.8 x 10-6) = 1.3 x 10-3

Note that, x = [H+] = 1.3 x 10-3 M; pH = - log(1.3 x 10-3) = 2.89 Percent Dissociation

The percent dissociation (or degree of ionization) of a weak acid is defined as

follows:

Percent dissociation = Amount of acid that dissociate (mol/L) Initial concentration of acid (mol/L)



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For strong acids, the percent dissociation at equilibrium is almost 100%. For weak acids, the percent dissociation depends on Ka of the acid and their initial concentration. For example, the percent dissociation of acetic acid (HC2H3O2, Ka = 1.8 x 10-5) at 0.10 M concentration is,

(1.3 x 10-3 M/0.10 M) x 100% = 1.3 %

Ka or pKa is a measure of the acid strength. The stronger the acid, the more of it ionizes, and the greater its Ka (or the lower its pKa). The percent dissociation of a weak acid then depends on its Ka and also on the degree of dilution. In the more dilute acid, a higher percentage of the molecules will ionize. Now consider a solution of 0.010 M acetic acid and its ionization products:

Concentration: HC3H3O2(aq) ⇋ H+(aq) + C2H3O2

-(aq)

Initial [ ], M: 0.010 0.00 0.00

Change, [ ], M: -x +x +x

Equilibrium [ ], M : (0.010 - x) x x The acid ionization constant, Ka, is given by the expression:

Ka = [H+][C2H3O2-] = x2 = 1.8 x 10-5

[HC2H3O2] (0.010 � x) Ka << 0.010, we can still use the approximation that x << 0.010, and (0.010 � x) ~ 0.010 Then, Ka = x2 ~ x2/0.010 = 1.8 x 10-5 ; (0.010 � x)

x2 = (0.010)(1.8 x 10-5); x = (1.8 x 10-7) = 4.2 x 10-4

Note that, x = [H+] = 4.2 x 10-4 M; pH = - log(4.2 x 10-4) = 3.37 The degree of ionization of acetic acid at this concentration is,

4.2 x 10-4 M x 100% = 4.2 % 0.010 M

Note that, the degree of ionization of the acid is higher in the more dilute acid solution. In fact, in 1.0 M acetic acid, the percent dissociation is only 0.42%, which is a 10-fold lower than in 0.010 M acid solution. Exercise: 1. What is [H+] in 0.10 M HNO2 solution and the pH of the solution? Ka = 4.0 x 10-4? (Answer: [H=] = 6.3 x 10-3 M; pH = 2.20) 2. A 0.10 M aqueous solution of lactic acid, HC3H5O3, has a pH = 2.43. Determine [H+] in

this solution and the Ka of lactic acid. (Answer: [H+] = 3.7 x 10-3 M; Ka = 1.4 x 10-4) 3. If a 0.10 M benzoic acid, HC7H5O2 is 2.5% ionized, what is [H+] and pH of 0.10 M

benzoic acid? Calculate the Ka of benzoic acid. (Answer: [H+] = 2.5 x 10-3 M; pH = 2.60; Ka = 6.4 x 10-5)

_________________________________________________________________________ Bases

Strong Bases: Hydroxides of Group IA metals (LiOH, NaOH, KOH, RbOH, and CsOH) are strong

bases, but only NaOH and KOH are commercially important and are common



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laboratory bases, since the others are quite expensive. These bases are quite soluble in water and they completely dissociate in aqueous solution, producing a high concentration of hydroxide ions. A moderately dilute solution of NaOH contains [OH-] = Initial [NaOH].

NaOH(s) Na+(aq) + OH-

(aq)

A 0.10 M solution of NaOH contains [OH-] = 0.10 M.

Among the hydroxides of the alkaline earth metals, Ca(OH)2, Sr(OH)2, and Ba(OH)2 are also relatively strong bases. When completely dissolved, these bases produce two moles of OH- per mole of the base:

Ba(OH)2(aq) Ba2+

(aq) + 2 OH-(aq)

A 0.10 M Ba(OH)2 solution contains 0.20 M of OH- ions.

The hydroxides of other metals are only slightly soluble. Even in a saturated solution, the concentration of OH- is still quite low. For example, a saturated solution of Mg(OH)2 has [OH-] ~ 3.1 x 10-4 M.

Weak Bases Ammonia is the only weak base that is of commercial importance. Ammonia is very soluble in water and ionizes as follows:

NH3(aq) + H2O(l) ⇋ NH4+

(aq) + OH-(aq)

The base dissociation constant, Kb, is given by the expression:

Kb = ]NH[ 3

]-OH][4

[NH

= 1.8 x 10-5

The concentration of OH- of a weak base solution, such as NH3(aq), depends on its Kb and the initial concentration of the base. For example, to determine [OH-], pOH and pH of 0.10 M NH3(aq), we can set up the following �ICE� table:

Concentration: NH3(aq) + H2O(l) ⇋ NH4+

(aq) + OH-(aq)

Initial [ ], M: 0.10 0.00 0.00

Change, [ ], M: -x + x + x

Equilibrium [ ], M : (0.10 - x) x x

Kb = [NH4+][OH-] = x2 = 1.8 x 10-5

[NH3] (0.10 � x)

Using approximation, Kb = x2 ~ x2/0.10 = 1.8 x 10-5 ; (0.10 � x)

x2 = (0.10)(1.8 x 10-5); x = (1.8 x 10-6) = 1.3 x 10-3

where, x = [OH-] = 1.3 x 10-3 M; pOH = - log(1.3 x 10-3) = 2.87; pH = 11.13

For ammonia solution at 0.10 M concentration, the percent dissociation is:

Percent dissociation = 1.3 x 10-3 M x 100% = 1.3% 0.10 M



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Like weak acids, the percent dissociation of weak bases depends on the Kb of the base and the degree of dilution � larger Kb and higher dilution higher percent dissociation. Exercise: 1. What is [OH-] in 0.0055 M KOH and what is the pH of the solution? 2. What is [OH-] in a 0.020 M solution of Ba(OH)2? What is the pH of this solution? 3. A 0.10 M solution of methylamine, CH3NH2 is 6.5% ionized at 25 oC. Determine the Kb

for methylamine? What is the pH of this solution? (Answer: Kb = 4.5 x 10-4; pH = 11.81)

CH3NH2(aq) + H2O(l) ⇋ CH3NH3+

(aq) + OH-(aq)

_____________________________________________________________________ Polyprotic Acids

Acids such as HCl, HF, HOCl, HNO2, HC2H3O2, and other similar type of acids are called monoprotic acids - each contains a single ionizable hydrogen ion. Some acids contain more than one ionizable hydrogen, and they are called polyprotic acids. Examples of polyprotic acids are H2SO4, H2SO3, H2C2O4, and H3PO4. In each case, the hydrogen ionizes in stages and with different ionization constant, such as the following example with H3PO4:

H3PO4(aq) ⇋ H+(aq) + H2PO4

-(aq); Ka1 = 7.5 x 10-3;

H2PO4-(aq) ⇋ H+

(aq) + HPO42-

(aq); Ka2 = 6.2 x 10-8;

HPO42-

(aq) ⇋ H+(aq) + PO4

3-(aq); Ka3 = 4.8 x 10-13;

Acid strength decreases in the order: H3PO4 >> H2PO4- >> HPO4

2-; Sulfuric acid is a strong acid because the first hydrogen ionizes completely:

H2SO4(aq) H+(aq) + HSO4-(aq); Ka1 = very large

However, the second hydrogen does not dissociate complete. Thus, HSO4- is a weak acid,

which dissociates as follows:

HSO4-(aq) ⇋ H+

(aq) + SO42-

(aq); Ka2 = 1.2 x 10-2 Exercise: 1. What is [H+], [HSO4

-], and [SO42-] in 0.10 M H2SO4(aq)? What is the pH of the

solution? (Ka1 is very large; Ka2 = 1.2 x 10-2) (Answer: pH = 0.96)

2. What is the pH of 0.10 M oxalic acid, H2C2O4, and what is [C2O42-] in this solution?

(Ka1 = 6.5 x 10-2; Ka2 = 6.1 x 10-5 ) (Answer: [H+] = 0.054 M; pH = 1.26; [C2O4

2-] = 6.1 x 10-5 M) _________________________________________________________________________

Acid-Base Properties of Salts

When salts (ionic compounds) dissolve in water, we assume that they completely dissociate into separate ions. Some of these ions can react with water and behave as acids or bases. Salts are also products of acid-base reactions. The acidic or basic nature of a salt solution depends on whether it is a product of:

a strong acid-strong base reaction; a weak acid-strong base reaction; a strong acid-weak base reaction, or a weak acid-weak base reaction.



12

Salts of Strong Acid-Strong Base Reactions: such as NaCl, NaNO3, KBr, etc.

This type of salt forms a solution that is neutral, because neither the cation ion nor the anion reacts with water and offset the equilibrium between the H3O

+ and OH- concentrations in the aqueous solution.

Salts of Weak Acid-Strong Base Reactions: such as NaF, NaNO2, NaC2H3O2, etc.

A salt that is the product of a reaction between a weak acid and a strong base will form an aqueous solution that is basic in nature.

the anion of such salt reacts with water to increase [OH-].

For example, Sodium acetate (NaC2H3O2) is a product of reaction between a weak acid (HC2H3O2) and a strong base (NaOH).

HC2H3O2(aq) + NaOH(aq) NaC2H3O2(aq) + H2O(l) When sodium acetate is dissolved in water, it dissociates into sodium and acetate ions:

NaC2H3O2(aq) Na+(aq) + C2H3O2

-(aq)

The acetate ion undergoes hydrolysis in aqueous solution as follows:

C2H3O2-(aq) + H2O(l) ⇋ HC2H3O2(aq) + OH-

(aq); Kb = [HC2H3O2][OH-] [C2H3O2

-] Note the for the dissociation of acetic acid:

HC2H3O2(aq) + H2O(l) ⇋ H3O+

(aq) + C2H3O2-(aq); Ka = [H3O

+][C2H3O2-]

[HC2H3O2]

Ka x Kb = [H3O+][C2H3O2

-] x [HC2H3O2][OH-] = [H3O+][OH-] = Kw = 1.0 x 10-14

[HC2H3O2] [C2H3O2-]

Thus, for C2H3O2-, Kb (for = Kw = 1.0 x 10-14 = 5.6 x 10-10

Ka (for HC2H3O2) 1.8 x 10-5

Thus, the hydrolysis of acetate ion has Kb = 5.6 x 10-10 (> Kw), which implies that the solution of sodium acetate will be basic (pH > 7). Salts of Strong Acid-Weak Base Reactions: NH4Cl, NH4NO3, (CH3)2NH2Cl, C5H5NHCl.

Aqueous solutions of this type of salts are acidic, because the cation (such as NH4+

ion) is the conjugate acid of a weak base (NH3) the cations react with water, which increases [H3O

+] in solutions. For example, ammonium chloride, NH4Cl, is produced by the reaction of a strong acid HCl and a weak base NH3.

HCl(aq) + NH3(aq) NH4Cl(aq) NH4+

(aq) + Cl-(aq) In an aqueous solution, ammonium ion (NH4

+) establishes the following equilibrium that increases [H3O

+] and makes the solution acidic:

NH4+

(aq) + H2O(l) ⇋ H3O+

(aq) + NH3(aq); Ka = [H3O+][NH3]

[NH4+]

For the reaction of NH3 with water, the following equilibrium occurs:

NH3(aq) + H2O(l) ⇋ NH4+

(aq) + OH-(aq); Kb = [NH4

+][OH-] [NH3]



13

Ka x Kb = [H3O+][NH3] x [NH4

+][OH-] = [H3O+][OH-] = Kw = 1.0 x 10-14

[NH4+] [NH3]

Thus, for NH4+, Ka = Kw = 1.0 x 10-14 = 5.6 x 10-10

Kb (for NH3) 1.8 x 10-5

Thus, the hydrolysis of ammonium ion has Ka = 5.6 x 10-10 (> Kw), which implies that aqueous solution of ammonium chloride will be acidic (pH < 7). Salts of Weak Acid-Weak Base Reactions: such as NH4C2H3O2, NH4CN, NH4NO2, etc..

Aqueous solutions of this type of salts can either be neutral, acidic, or basic, depending on the relative magnitude of the Ka of the weak acid and the Kb of the weak base.

If Ka ~ Kb, the salt solution will be approximately neutral; If Ka > Kb, the salt solution will be acidic, and If Ka < Kb, the salt solution will be basic.

For example, Ka of HC2H3O2 = 1.8 x 10-5, Kb of NH3 = 1.8 x 10-5, and aqueous solution of ammonium acetate, NH4C2H3O2, is approximately neutral.

NH4C2H3O2(aq) NH4+

(aq) + C2H3CO2-(aq)

NH4+

(aq) + H2O(l) ⇋ H3O+

(aq) + NH3(aq); Ka = 5.6 x 10-10;


(aq); Kb = 5.6 x 10-10;

Ka=Kb a neutral solution, because the hydrolyses result in a solution with [H3O+] = [OH-].

For NH4CN, Ka(HCN) = 6.2 x 10-10, Kb(NH3) = 1.8 x 10-5, and aqueous solution of NH4CN will be basic. NH4CN(aq) NH4

+(aq) + CN-

(aq)

NH4+

(aq) + H2O(l) ⇋ H3O+

(aq) + NH3(aq); Ka = 5.6 x 10-10;

CN-(aq) + H2O(l) ⇋ HCN(aq) + OH-

(aq); Kb = 1.6 x 10-5; Ka < Kb basic solution, because the hydrolyses result in a solution in which [H3O

+] < [OH-]. For NH4NO2, Ka(HNO2) = 4.0 x 10-4, Kb(NH3) = 1.8 x 10-5, and aqueous solution of NH4NO2 will be acidic. NH4NO2(aq) NH4

+(aq) + NO2

-(aq)

NH4+

(aq) + H2O(l) ⇋ H3O+

(aq) + NH3(aq); Ka = 5.6 x 10-10;

NO2-(aq) + H2O(l) ⇋ HNO2(aq) + OH-

(aq); Kb = 2.5 x 10-11;

Ka > Kb acidic solution, because the hydrolyses result in a solution with [H3O+] > [OH-].

Calculating the pH of a Basic or an Acidic salt Solution

1. Consider a solution of 0.050 M sodium acetate, which dissociates completely as follows:

NaC2H3O2(aq) Na+(aq) + C2H3O2

-(aq)

The acetate ion undergoes hydrolysis in aqueous solution as follows:


(aq); Kb = [HC2H3O2][OH-] = 5.6 x 10-10 [C2H3O2

-]

By approximation, [OH-] = (Kb[C2H3O2-]) = {(5.6 x 10-10)(0.050)} = 5.3 x 10-6 M

pOH = -log(5.3 x 10-6) = 5.28, and pH = 8.72, ( a basic solution)



14

2. Consider a solution of 0.050 M NH4Cl, which dissociates as follows:

NH4Cl(aq) NH4+

(aq) + Cl-(aq)

In an aqueous solution, the following equilibrium occurs:

NH4+

(aq) + H2O(l) ⇋ H3O+

(aq) + NH3(aq); Ka = [H3O+][NH3] = 5.6 x 10-10

[NH4+]

By approximation, [H3O+] = (Ka[NH4

+]) = {(5.6 x 10-10)(0.050)} = 5.3 x 10-6 M

pH = -log(5.3 x 10-6) = 5.28, ( an acidic solution) Exercise: 1. Predict whether aqueous solution of each of the following salts is acidic, basic, or

neutral?

(a) KNO3 (b) CH3NH3Cl (c) Na2SO3 (d) (NH4)2HPO4

Ka (for HSO3-) = 1.0 x 10-7; Ka (for H2PO4

-) = 6.2 x 10-8; Kb (for CH3NH2) = 4.4 x 10-4 2. Calculate the pH of a solution that is 0.10 M in sodium cyanide, NaCN(aq). (Ka for HCN = 6.2 x 10-10) (Answer: pH = 11.10) 3. What is the pH of a solution containing 1.5 g of pyridinium chloride, C5H5NHCl, in

100.0 mL solution? (Kb of C5H5N = 1.7 x 10-9) In aqueous solution C5H5NHCl dissociates as follows:

C5H5NHCl(aq) C5H5NH+(aq) + Cl-(aq) (Answer: pH = 3.56)

________________________________________________________________________

The Common Ion Effect in Acid-Base Equilibria Common ions are ions produced by more than one solute in the same solution. For example, in a solution containing sodium acetate and acetic acid, the acetate ion (C2H3O2

-) is the common ion. According to the Le Chatelier's principle, the following equilibrium for acetic acid:

HC2H3O2(aq) + H2O(l) ⇌ H3O+

(aq) + C2H3O2-(aq)

will shift to the left if C2H3O2

- (from another source) is added to the solution. This will reduce the acid dissociation, which lowers [H3O

+] and increases the pH of the solution. Addition of ammonium chloride, NH4Cl, to ammonia solution will cause the following equilibrium to shift to the left:

NH3(aq) + H2O(l) ⇌ NH4+

(aq) + OH-(aq)

NH4Cl dissociates to produce NH4+ and Cl- ions, where NH4

+ is a common ion in the equilibrium of ammonia solution. The introduction of NH4

+ ion reduces the ionization of NH3, which decreases [OH-] and lowers the pH of the solution. Calculating [H+] and pH of a solution containing weak acid and the salt of its conjugate base.

Consider a solution containing 0.10 M HNO2 and 0.050 M NaNO2, where the latter dissociates completely as follows: NaNO2(aq) Na+

(aq) + NO2-(aq)

The concentration of H+ can be calculated using the following �ICE� table:



15

Concentration: HNO2(aq) ⇌ H+(aq) + NO2

-(aq)

Initial [ ], M 0.10 0.00 0.050 (from salt)

Change, [ ], M -x +x +x

Equilibrium [ ], M (0.10 � x) x (0.050 + x)

Ka = ][HNO

]][NO[H

2

-2

= ) - (0.10

) )(0.050(

x

xx = 4.0 x 10-4

Since Ka << 0.10, x is assumed to very small, and by approximation,

) - 0.10(

) )(0.050(

x

xx ~

(0.10)

)(0.050)(x = 4.0 x 10-4; x = 8.0 x 10-4

where x = [H+] = 8.0 x 10-4 M, and pH = -log(8.0 x 10-4) = 3.10 In general, for a solution containing a weak acid HX, with an dissociation constant Ka, and the salt NaX, [H+] can be calculated by approximation, such that,

[H+] = Ka x ][

][X

HX

Exercises: 1. What are [H+] and pH of a solution containing 4.0 g of sodium acetate (NaC2H3O2) in

1.0 L of 0.050 M HC2H3O2(aq)? (Ka = 1.8 x 10-5 for acetic acid) 2. What are [OH-], [H+], and pH of a solution containing 2.5 g of NH4Cl dissolved in 1.0 L

of 0.050 M NH3(aq)? (Kb = 1.8 x 10-5 for NH3) Buffered Solutions

A buffered solution is one that has the ability to maintain its pH relatively constant even when a small amount of strong acid or strong base is added.

Buffered solutions contain a comparable amount of weak acid and its conjugate base (or a weak base and its conjugate acid).

Consider a buffer made up of acetic acid and sodium acetate, in which the major species present in solution are HC2H3O2 and C2H3O2

-.

If a little HCl is added to this solution, most of H+ (from HCl) is absorbed by the conjugate base, C2H3O2

-, in the following reaction:

H+(aq) + C2H3O2

-(aq) HC2H3O2

-(aq)

Since C2H3O2

- is present in a much larger quantity than the added H+, the above reaction is considered to go almost completely to the right. This buffering reaction prevents a significant increase in [H+] and minimizes the change in its pH. If a strong base such as NaOH is added, most of the OH- ions (from NaOH) are reacted by the acid component of the buffer as follows:

OH-(aq) + HC2H3O2(aq) H2O + C2H3O2

-(aq).

Again, because of the larger concentration of HC2H3O2 compared to OH-, this reaction also

goes almost to completion. This buffering reaction prevents a large increase in the [OH-] and minimizes any change in the pH of the solution.



16

Buffer solution is generally prepared by dissolving a mixture of either a weak acid and a "salt" that contains its conjugate base, or a weak base and a "salt" that contains its conjugate acid, in the desired molar ratio to obtain the correct pH. A given buffer is effective within a range of pH that is typically within approximately 1 of the pKa of its acid component. Examples of some common buffer systems:

Buffer pKa pH Range_____

HCHO2 � NaCHO2 3.74 2.75 � 4.75

HC2H3O2 � NaC2H3O2 4.74 3.75 � 5.75

KH2PO4 - K2HPO4 7.21 6.20 � 8.20 (one of the buffer systems in blood);

CO2/H2O � NaHCO3 6.37 5.40 � 7.40 (one of the buffer systems in blood)

NH3 - NH4Cl 9.25 8.25 � 10.25 Calculating The [H+] and pH of Buffer Solutions:

For a buffer containing the weak acid HB and the salt NaB, such that B- is the conjugate base to the acid, the concentration [H+] and pH of the buffer depend on the dissociation constant, Ka, of the acid component and the concentration ratio [B-]/[HB] in the buffer solution.

Consider the equilibrium: HX(aq) ⇌ H+(aq) + X-

(aq); Ka = ][

]][[

HX

XH

Re-arranging, we obtain, [H+] = Ka x ][

][X

HX; pH = pKa + log(

][

][X

HX)

The last expression is called the Henderson-Hasselbalch equation, which is useful for calculation the pH of solutions when the ratio [X-]/[HX] is known. For a given buffer system, [H+] and pH varies with the ratio: [Base]/[Acid]. Exercise: 1. A 100.0 mL buffer solution contains 5.0 g of 0.038 mol acetic acid and 0.061 mol of

sodium acetate. Calculate the pH of the buffer? (Ka = 1.8 x 10-5 for acetic acid) If 0.0050 mol of HCl is added to the buffer, what is the final pH of the solution?

(Assume that the volume of buffer does not change when HCl(aq) is added.) 2. A buffer solution is prepared by mixing 25 mL of 0.20 M H2SO3 and 75 mL of 0.20 M

NaHSO3. Calculate the pH of this buffer. (Ka = 1.5 x 10-2 for H2SO3) 3. (a) Suppose you want to prepare a 100-mL buffer solution with pH = 7.30, which

buffer system would you consider? [Hint: look for acids that have pKa close to the buffer pH.] (b) Calculate the ratio [Base]/[Acid] that the buffer should have. (c) If the concentration of the acid is 0.50 M, what should be the base concentration? (d) Assuming your starting materials are solids, how many grams of each must be dissolved to make the 100. mL buffer solution? (Both acid and its conjugate base may be in the form of sodium or potassium salts)

Buffering Capacity

The buffering capacity of a solution represents the amount of H+ and/or OH- the buffer solution can absorb without significantly altering its pH. A buffer that contains large concentrations of buffering components and able to absorb significant quantities of strong acid or strong base with little change in its pH is said to have a large buffering capacity.



17

Whereas the pH of a buffered solution is determined by the ratio [X-]/[HX], buffer capacity is determined by the sizes of [HX] and [X-] in the buffer. Calculating the change in pH of buffered solution after a strong acid is added:

1. Calculate the change in pH that occurs when 0.010 mol of HCl is added to 1.0 L of each of the following buffers:

Buffer-A: 1.0 M HC2H3O2 + 1.0 M NaC2H3O2;

Buffer-B: 0.020 M HC2H3O2 + 0.020 M NaC2H3O2 Solutions

For these buffers the pH can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([C2H3O2-]/[HC2H3O2]) = -log(1.8 x 10-5) + log(1) = 4.74

When 0.010 mol HCl is added to Buffer-A, the following reaction takes place:

H+(aq) + C2H3O2

-(aq) ( HC2H3O2(aq)

[ ] before reaction: 0.010 M 1.0 M 1.0 M

[ ] after reaction: 0 0.99 M 1.01 M

The new pH = 4.74 + log(0.99/1.01) = 4.74 - 0.010 = 4.73 (pH is hardly changed ~ 0.21%) When 0.010 mol HCl is added to Buffer-B, the following reaction takes place:

H+(aq) + C2H3O2-(aq) ( HC2H3O2(aq) [ ] before reaction: 0.010 M 0.020 M 0.020 M

[ ] after reaction: 0 0.010 M 0.030 M New pH = 4.74 + log(0.010/0.030) = 4.74 � 0.48 = 4.26 (pH decreases by 0.48 unit or 10%)

This shows that Buffer-A, which contains larger quantities of buffering components, has a much higher buffering capacity that Buffer-B. For the Buffer-A to decrease its pH by 0.48 unit (or 10%), it would have to absorb the equivalent of 0.50 mol of HCl. Exercise 1. An acetate buffer solution is prepared by mixing 35.0 mL of 1.0 M HC2H3O2 and 65.0 mL

of 1.0 M NaC2H3O2. Calculate the pH of this buffer. If 1.0 mL of 6.0 M HCl is added to the buffer solution, what will be the new pH of the buffer.

2. A 100-mL buffer solution contains 0.20 M KH2PO4 and 0.32 M K2HPO4. What is the pH

of the buffer? If 0.0050 mol of HCl is added to the buffer without changing its volume, what is the final pH of the buffer? (Ka = 6.2 x 10-8 for H2PO4

-) _________________________________________________________________________

Titration and pH Curves

A pH curve is a graph of pH versus the volume of titrant in an acid-base titration. The data used to plot a pH curve may be obtained either by computation or by measuring the pH directly with a pH meter during titration. There are four types of pH curves for the different types of acid-base titration:

1. strong acid - strong base titration. 2. strong acid - weak base titration.



18

3. weak acid - strong base titration. 4. weak acid - weak base titration.

(MEMORIZE AND UNDERSTAND THE GENERAL SHAPES AND FEATURES OF EACH pH CURVES!) Strong Acid-Strong Base Titration

Net reaction: H+(aq) + OH-

(aq) H2O(l) Calculating The pH of Solutions During Titration:

Consider the titration of 20.0 mL of 0.100 M HCl with 0.100 M NaOH solution. Calculate the pH of the solution: (a) before any of the NaOH is added; (b) after 15.0 mL of NaOH is added; (c) after 19.5 mL of NaOH is added; (d) after 20.0 mL of NaOH is added; (e) after 21.0 mL of NaOH is added. (a) Before titration, [H+] = 0.100 M, pH = 1.000 (b) After 15.0 mL of NaOH is added, calculation of [H+] will be as follows:

H+(aq) + OH-

(aq) H2O [ ] before mixing: 0.100 M 0.100 M [ ] after mixing, but before reaction: 0.057 M 0.043 M [ ] after reaction: 0.014 M 0.000

[H+] = 0.014 M, pH = -log(0.014) = 1.85 (c) After 19.0 mL of NaOH is added, calculation of [H+] will be as follows:

H+(aq) + OH-

(aq) H2O [ ] before mixing: 0.100 M 0.100 M [ ] after mixing, but before reaction: 0.051 M 0.049 M [ ] after reaction: 0.002 M 0.000

[H+] = 0.002 M, pH = -log(0.002) = 2.70 (d) After 20.0 mL of NaOH is added, calculation of [H+] will be as follows:

H+(aq) + OH-

(aq) ( H2O [ ] before mixing: 0.100 M 0.100 M [ ] after mixing, but before reaction: 0.050 M 0.050 M [ ] after reaction: 0.000 M 0.000 At this point, the solution contains only water and salt (NaCl) and [H+] = [OH-] = 1.0 x 10-7 M, which is due to water ionization. The pH of the solution will be 7.00 (pH of neutral solution) (e) When 21.0 mL of 0.100 M NaOH has been added, there will be excess OH-:

H+(aq) + OH-(aq) ( H2O [ ] after mixing, but before reaction: 0.049 M 0.051 M [ ] after reaction: 0.000 0.002 M [OH-] = 0.002 M, ( pOH = 2.70, and pH = 11.30



19

Note that in strong acid-strong base titration, an abrupt change from about pH 3 to 11 occurs within (0.5 mL of NaOH added near the equivalent point. All strong acid-strong base titration have the same pH curves. Weak Acid-Strong Base Titration

Net reaction: HX(aq) + OH-(aq) ( H2O + X-(aq)

For example, when acetic acid is titrated with sodium hydroxide, the net reaction is

HC2H3O2(aq) + OH-(aq) ( C2H3O2-(aq) + H2O 1. Consider the titration of 20.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH solution.

Calculate the pH of the solution: (a) before any of the NaOH is added; (b) after 10.0 mL of NaOH is added; (c) after 15.0 mL of NaOH is added; (d) after 20.0 mL of NaOH is added; (e) after 25.0 mL of NaOH is added.

Solution:

(a) Before titration, [H+] = {(0.100 M)(1.8 x 10-5)} = 1.34 x 10-3 M,

pH = -log(1.34 x 10-3) = 2.873 (b) After adding 10.0 mL of 0.100 M NaOH, [H+] is calculated as follows:

Concentration, M: HC2H3O2(aq) + OH-(aq) C2H3O2

-(aq) + H2O

Before mixing: 0.100 0.100 0.000

After mixing, before rxn: 0.0667 0.0333 0.000

After reaction: 0.0333 0.000 0.0333 Using Henderson-Hasselbalch equation,

[H+] = Ka x ]OH[C

]OH[HC-232

232 = 1.8 x 10-5 x M) (0.033

M) 033.0( = 1.8 x 10-5 M

pH = pKa + log(]OH[C

]OH[C

232

-232 = 4.74 + log(1) = 4.74

When a weak acid is half-neutralized, 50% of the acid is converted to its conjugate base. That is, at half-way to the equivalent point of the titration, [C2H3O2

-] = [HC2H3O2]. Under this condition, [H+] = Ka, and pH = pKa. (c) At the equivalent point (when 20.0 mL of 0.100 M NaOH has been added), all of the acid is converted to its conjugate base. The latter undergoes hydrolysis as follows:

C2H3O2-(aq) + H2O(l) ⇌ HC2H3O2(aq) + OH-

(aq) �� Initial [ ], M: 0.0500 0.000 0.000 Change, [ ], M: -x +x +x Equilibrium [ ], M: (0.0500 � x) x x

By approximation, [OH-] = x = 0-232 ]OH[C x bK = )(0.050)10 x 5.6( -10 = 5.3 x 10-6 M

pOH = -log(5.3 x10-6) = 5.28; pH = 8.72



20

The hydrolysi of the conjugate base of a weak acid produces OH-, which leads to pH > 7 at the equivalent point. In the case of acetic-NaOH titration, the pH at the equivalent point is about 8.72. The larger the Ka of the weak acid, the closer the pH to 7 at equivalent point. The smaller Ka, the higher the pH than 7 at equivalent point. Exercise 1. In a titration of 20.00 mL of 0.100 M formic acid, HCHO2, with 0.100 M NaOH,

calculate the pH: (a) before NaOH was added; (b) after 15.00 mL of NaOH is added; (b) after 20.00 mL of NaOH is added. (Ka = 1.8 x 10-4 for HCHO2)

2. In a titration of 20.00 mL of 0.100 M NH3(aq) with 0.100 M HCl(aq), calculate the pH: (a) before titration begins; (b) after 10.0 mL of HCl(aq) is added; (c) at equivalent point, that is after 20.0 mL of HCl(aq) is added. (Kb = 1.8 x 10-5 for NH3)

_________________________________________________________________________

The Use of Indicators in Acid-Base Titration

An indicator - a substance that changes color to mark the end-point of a titration. Most indicators used in acid-base titration are weak organic acids. Indicators exhibit one color in the acid or protonated form (HIn) and another in the

base or deprotonated form (In-). Each indicator has a range of pH = pKa 1, where the change of colors occurs; A suitable indicator gives the end-point that corresponds to the equivalent point of

the titration; this would be one with a range of pH that falls within the sharp increase (or decrease) of pH in the titration curves.

Like a weak acid, an indicator exhibits the following equilibrium in aqueous solution:

HIn(aq) ⇌ H+(aq) + In-

(aq);

Ka = [HIn]

]][In[H -

;

Re-arranging, we obtain [H+] = Ka x ([HIn]/[In-]); pH = pKa + log([HIn]

]In[ -

)

When [HIn] = 10 x [In-],

pH = pKa + log([HIn]

]In[ -

) = pKa - 1; indicator assumes the color of acid form

When [In-] = 10 x [HIn],

pH = pKa + log([HIn]

]In[ -

) = pKa + 1; indicator assumes the color of base.

Phenolphthalein, which is the most common acid-base indicator, has Ka ~ 10-9. Its acid form (HIn) is colorless and the conjugate base form (In-) is pink. It is colorless when the solution pH < 8, that is when 90% or more of the species are in the acid form (HIn), and pink at pH > 10, when 90% or more of the species are in the conjugate base form (In-). The pH range at which an indicator changes depends on the Ka. For phenolphthalein with Ka ~ 10-9, its color changes at pH = 8 � 10. It is suitable for strong acid-strong base titration and weak



21

acid-strong base titration. The range of pH for other common acid-base indicators are listed below: Indicators: Acid Color Base Color pH Range Type of Titration Methyl orange orange yellow 3.2 � 4.5 strong acid-strong base strong acid-weak base

Bromocresol green yellow blue 3.8 � 5.4 strong acid-strong base strong acid-weak base

Methyl red Red Yellow 4.5 - 6.0 strong acid-strong base strong acid-weak base

Bromothymol blue Yellow blue 6.0 - 7.6 strong acid-strong base

Phenol Red orange red 6.8 � 8.2 strong acid-strong base weak acid-strong base

Solubility Equilibria Solubility Equilibria and the Solubility Product Constants

When a slightly soluble salt such as silver chloride, AgCl, is dissolved in water, a saturated is quickly obtained, because only a very small amount of the solid dissolves, while the rest remains undissolved. The following equilibrium between solid AgCl and the free ions occurs in solution: AgCl(s) ⇌ Ag+(aq) + Cl-(aq);

Ksp = [Ag+][Cl-] Ksp is called the solubility product constant or just solubility product. General Expressions of Solubility Product Constant, Ksp

For solubility equilibrium: MaXb(s) ⇌ aMb+(aq) + bXa-

(aq),

Ksp = [Mb+]a[Xa-]b

A. For ionic equilibria of the type: MX(s) ⇌ Mn+(aq) + Xn-

(aq); Ksp = [Mn+][Xn-]

If the solubility of compounds is S mol/L, Ksp = S2; S = (Ksp)

For example, the solubility equilibrium for BaSO4 is

BaSO4(s) ⇌ Ba2+(aq) + SO4

2-(aq); Ksp = [Ba2+][SO4

2-] = 1.5 x 10-9

If the solubility of BaSO4 is S mol/L, a saturated solution of BaSO4 has

[Ba2+] = [SO42-] = S mol/L, Ksp = S2; and S = (Ksp) = (1.5 x 10-9) = 3.9 x 10-5

mol/L

B. For ionic equilibria of the type: MX2(s) ⇌ M2+(aq) + 2X-

(aq), Ksp = [M2+][X-]2 ;

and for the type: M2X(s) ⇌ 2 M+(aq) + X2-

(aq); Ksp = [M+]2[X2-]

For both types, if the solubility is S mol/L, Ksp = 4S3; S = (Ksp/4)1/3

For example, CaF2(s) ⇌ Ca2+(aq) + 2F-

(aq); Ksp = [Ca2+][F-]2 = 4.0 x 10-11

The solubility of calcium fluoride is S = (Ksp/4)1/3 = (4.0 x 10-11/4)1/3 = 2.2 x 10-4 mol/L



22

C. For solubility equilibria of the type: MX3(s) ⇌ M3+(aq) + 3X-

(aq), Ksp = [M3+][X-]3

Or one of the type: M3X(s) ⇌ 3 M+(aq) + X3-

(aq), Ksp = [M+]3[X3-]

If the solubility of the compound (MX3 or M3X) is S mol/L, Ksp = 27S4; S = 4 )27/( spK

For example, Ag3PO4(s) ⇌ 3Ag+(aq) + PO4

3-(aq); Ksp = [Ag+]3[PO4

3-] = 1.8 x 10-18

The solubility of silver phosphate is

S = 4 )27/( spK = 4 -18 )/2710 x (1.8 = 1.6 x 10-5 mol/L

D. For solubility equilibria: M2X3(s) ⇌ 2M3+(aq) + 3X2-

(aq), Ksp = [M3+]2[X2-]3

OR, one of the type: M3X2(s) ⇌ 3M2+(aq) + 2X3-

(aq), Ksp = [M2+]3[X3-]2

If the solubility of the compound M3X2 is S mol/L, then [M2+] = 3S, and [X3-] = 2S;

Ksp = (3S)3(2S)2 = 108S5; S = 5 108/spK

For example, Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO4

3-(aq),

Ksp = [Ca2+]3[PO43-]2 = 1.3 x 10-32;

the solubility of Ca3(PO4)2 is S = 5 -32 )/10810 x (1.3 = 1.6 x 10-7 mol/L

Calculating Ksp from Solubility

1. Suppose the solubility of PbSO4 in water is 4.3 x 10-3 g/100 mL solution at 25 oC. What is the Ksp of PbSO4 at 25 oC?

Solution: Solubility of PbSO4 in mol/L = 4.3 x 10-3 g x 1000 mL/L = 1.40 x 10-4 mol/L 100. mL 303.26 g/mol

That is, a saturated solution of PbSO4, contains [Pb2+] = [SO42-] = 1.4 x 10-4 mol/L

For the equilibrium: PbSO4(s) ⇌ Pb2+(aq) + SO4

2-(aq);

Ksp = [Pb2+][SO42-] = S2 = (1.4 x 10-4 mol/L)2 = 2.0 x 10-8

Exercise 1. If the solubility of MgF2 is 7.3 x 10-3 g/100 mL solution at 25oC, what is the Ksp of

MgF2? 2. A saturated solution of calcium hydroxide has a pH = 12.17. Calculate the solubility of

Ca(OH)2 and its Ksp at 25oC. _________________________________________________________________________ Determining Solubility from Ksp



23

1. If the Ksp of Mg(OH)2 is 6.3 x 10-10 at 25 oC, what is its solubility at 25 oC (a) in mol/L, and (b) in g/100 mL solution at 25 oC?

(a) Solubility equilibrium for Mg(OH)2 is: Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH-

(aq)

Ksp = [Mg2+][OH-]2 = 4S3 = 6.3 x 10-10

Solubility of Mg(OH)2: S = 3 4/spK = 3 -10 )/410 x 6.3( = 5.4 x 10-4 mol/L

(b) Solubility in g/100 mL solution = (5.4 x 10-4 mol/L)(58.32 g/mol)(0.1L/100 mL)

= 3.1 x 10-3 g/100 mL solution. Exercise: 1. The Ksp of PbI2 is 1.4 x 10-8. What is its solubility in grams per 100. mL of solution at

25oC? 2. Calcium hydroxide, Ca(OH)2, has Ksp = 1.3 x 10-6. How many grams of Ca(OH)2 can be

dissolved in 250-mL solution to make a saturated solution at 25oC? _________________________________________________________________________ Relative Solubility from Ksp

For compounds whose formula yields the same number of ions, their Ksp value can be used to determine relative solubility. That is, larger Ksp implies greater solubility. Common Ion Effect of Solubility The presence of common ion decreases the solubility of a slightly soluble ionic compound. For example, in the following equilibrium

PbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq)

If some NaCl is added to a saturated solution of PbCl2, the [Cl-] will increase, and according to Le Chateliler�s principle, the equilibrium will shift in the direction that tends to reduce [Cl-

]. In this case, the equilibrium will shift left, to form more PbCl2 solid, hence decreasing the amount of PbCl2 that dissolves into solution. Sample Problem

The Ksp of BaSO4 is 1.5 x 10-9 at 25 oC. (a) What is its solubility in water at 25oC? (b) What is the solubility in 0.10 M Na2SO4 at 25 oC?

Solubility equilibrium for BaSO4: BaSO4(s) ⇌ Ba2+(aq) + SO4

2-(aq);

Ksp = [Ba2+][SO42-] = S2 = 1.5 x 10-9

Solubility of BaSO4 in water is S = (Ksp) = (1.1 x 10-10) = 3.9 x 10-5 mol/L

Let the solubility of BaSO4 in 0.10 M Na2SO4 be x mol/L

Then, the solution contains [Ba2+] = x mol/L and [SO42-] = (0.10 + x) mol/L

Ksp = [Ba2+][SO42-] = (x)(0.10 + x) = 1.5 x 10-9

By approximation, since x << 0.10 M, (0.10 + x) ~ 0.10, and

Ksp = (x)(0.10 + x) ~ (x)(0.10) = 1.5 x 10-9; x = 1.5 x 10-8 mol/L The Effect of pH on Solubility

pH affect the solubility of slightly soluble compounds containing anions that are conjugate bases of weak acids, such as F-, NO2

-, OH-, SO32-, and PO4

3-, but not those containing anions which are conjugate bases of strong acids, such as SO4

2-, Cl-, Br-. For example, in a saturated solution of calcium fluoride, CaF2, the following equilibrium exists:



24

CaF2(s) ⇌ Ca2+(aq) + 2F-

(aq)

When a strong acid is added to the saturated solution, the following reaction occurs:

H+(aq) + F-

(aq) HF(aq)

This reaction has the net effect of reducing the concentration of F- ions, which causes the equilibrium to shift to the right and resulting in more of CaF2 to dissolve. Exercise 1. Calculate the solubility of PbCl2 (Ksp = 1.6 x 10-5) and AgCl (Ksp = 1.6 x 10-10) in 0.010

M NaCl solution. 2. What is the solubility of CuS (Ksp = 4 x 10-37) and ZnS (Ksp = 2 x 10-25), respectively,

in solution buffered at pH = 8.00? (For H2S, Ka1 = 1 x 10-7; Ka2 = 1 x 10-19) _________________________________________________________________________ Precipitation and Qualitative Analysis

For solubility equilibrium: MaXb(s) ⇌ aM+(aq) + bX-

(aq),

We can express the ion product, Q, as follows: Q = [M+]oa[X-]o

b

Where [ ]o is the initial concentration of each ion in solution.

For a given pair of ions pertaining to a slightly soluble compound, if:

Q = Ksp; solution is saturated, but no precipitate is formed.

Q > Ksp; solution is saturated and a precipitate is formed.

Q < Ksp; solution is not saturated

Sample problem: If 20.0 mL of 0.050 M Pb(NO3)2 is mixed with 30.0 mL of 0.10 M NaCl, will PbCl2 precipitate form? Ksp of PbCl2 = 1.6 x 10-5.

Pb2+(aq) + 2Cl-(aq) ⇌ PbCl2(s)

[ ] after mixing 0.020 M 0.060 M

Qsp = [Pb2+][Cl-]2 = (0.020)(0.060)2 = 7.2 x 10-5 > Ksp

Qsp > Ksp PbCl2 ppt will form.

Exercise: 1. At what pH a solution containg 0.10 M Ca2+ ions will form a precipitate of Ca(OH)2?

(Ksp = 1.3 x 10-6 at 25 oC for Ca(OH)2) 2. Will Ag2SO4 precipitate if 30.0 mL of 0.050 M AgNO3 is added to 20.0 mL of 0.10 M of

Na2SO4 solution? Ksp = 1.2 x 10-5 for Ag2SO4 at 25oC. _________________________________________________________________________ Practical Application of Ionic Equilibria

Qualitative Analysis

The presence of certain ions, such as Cl-, I-, SO42-, Ag+, etc. in water can be

determined qualitatively and quantitatively by precipitation method. Cl- and I- can be precipitated as AgCl and AgI, respectively, by adding AgNO3 solution. Synthesis

Certain industrial chemicals, such as AgCl, AgBr, AgI, that are needed in photographic industries are prepared by precipitation.



25

AgNO3(aq) + NaBr(aq) AgBr(s) + NaNO3(aq)

Since the Ksp of these compounds are very small, the precipitation reaction practically goes to completion. The separation and purification processes are simple. Selective Precipitation

Because of the different solubility of compounds, the separation of ions from a mixture of solution can be accomplished by selectively precipitation. This selective precipitation is carried out by adding the precipitating ion until the ion product, Q, of the more soluble compound is almost equal to its Ksp, but the Ksp of the less soluble compound is exceeded as much as possible. For example, because of the different solubility of CaSO4 (Ksp = 6.1 x 10-5), and BaSO4 (Ksp = 1.5 x 10-9), Ca2+ and Ba2+ can be separated from a solution by selectively precipitating BaSO4 using Na2SO4 or H2SO4 solution. Sample problem-3: An aqueous solution contains Ca2+ and Ba2+ ions at concentrations 0.10 M and 0.010 M, respectively. What concentration of SO4

2- must be present so that the maximum amount of Ba2+ ion precipitates as BaSO4, but Ca2+ remains in solution ? Ksp of CaSO4 = 6.1 x 10-5; Ksp of BaSO4 = 1.5 x 10-9. Exercise-9: 1. How many grams of AgCl will precipitate out when 50.0 mL of 0.050 M AgNO3 is added

to 50.0 mL of 0.10 M NaCl? What is the concentration of Ag+ that remains in solution after the precipitation? Ksp = 1.6 x 10-10 at 25 oC for AgCl.

2. A solution contains 0.10 M in Cl- and 0.010 M in I-. At what concentration of Ag+ would (a) AgI begin to precipitate; (b) AgCl begins to precipitate? What is the concentration of I- when AgCl begins to precipitate? (Ksp[AgCl] = 1.6 x 10-10 and Ksp[AgI] = 1.5 x 10-16)

3. Sea water contains Mg2+ at a concentration of about 0.015 M. Would precipitate of Mg(OH)2 form if a sample of sea water is added to a saturated solution of Ca(OH)2 ? What concentrion of Mg2+ is present when the solution is just saturated with Ca(OH)2 ?

(Ksp = 1.3 x 10-6 for Ca(OH)2; Ksp = 8.9 x 10-12. Qualitative Analysis: Identifying Ions in Mixtures of Cations The separation and identification of ions in a mixture is called qualitative analysis. The technique employs the different solubility of ionic compounds in aqueous solution as well as the ability of certain cations to form complex ion with ligands. For many transition metals, their complex ions are often colored, which can be used in their identification. Separation into Groups

The general approach in the qualitative analysis of cations is to separate them into various ion groups. Ion group 1: Insoluble chlorides. Treating the mixture with 6 M HCl will precipitate Ag+, Hg2

2+, and Pb2+ ions as chlorides, leaving other cations in solution. The formation of a white precipitate indicates the presence of at least one of these cations in the mixture. Ion group 2: Acid-insoluble sulfides. The supernatant from the above treatment with HCl is adjusted to pH ~ 0.5 and then treated with aqueous H2S. The high [H3O

+] in solution keeps



26

[HS-] very low, which precipitates only the following group of cations: Cu2+, Cd2+, Hg2+, Sn2+, and Bi3+. Centrifuging and decanting gives the next solution. Ion group 3: Base-insoluble sulfides. The supernatant from acidic sulfide treatment is treated with NH3/NH4

+ buffer to make the solution slightly basic (pH~8). The excess OH- in solution increases [HS-], which causes the precipitation of the more soluble sulfides and some hydroxides. The cations that precipitate under this condition are: Zn2+, Mn2+, Ni2+, Fe2+, Co2+, as sulfides, and Al3+, Cr3+, and Fe3+ as hydroxides. The precipitate is centrifuged and the supernatant decanted to give the next solution. Ion group 4: Insoluble phosphates. The slightly basic supernatant separated from the group 3 ions is treated with (NH4)2HPO4, which precipitates Mg3(PO4)2, Ca3(PO4)2, and Ba3(PO4)2. Ion group 5: Alkali metal and ammonium ions. The final solution contains any of the following ions: Na+, K+, and NH4

+. Sample qualitative analysis. A solution contains a mixture of Ag+, Al3+, Cu2+, and Fe3+ ions. Devise a scheme to separate and identify each of these cations. Mixture: Ag+, Al3+, Cu2+, Fe3+ (colored solution)

add 3 M HCl, centrifuge Precipitate (white) Supernatant (colored) AgCl (Al3+, Cu2+, Fe3+) add 6 M NaOH, centrifuge Precipitates Supernatant (colorless) (Cu(OH)2 & Fe(OH)3) Al(OH)4

-

add 6 M NH3 Precipitate Supernatant Fe(OH)3 Cu(NH3)4

2+ (dark brown) (dark blue) ______________________________________________________________________________ Complex Ion Equilibria 15.8 Equilibria Involving Complex Ions



27

A complex ion consists of a central metal ion that is covalently bonded to two or more ligands, which can be anions such as OH-, Cl-, F-, CN-, etc. or neutral molecules such as H2O, CO, and NH3. For example, in the complex ion [Cu(NH3)4]2+, Cu2+ is the central metal ion with four NH3 molecules covalently bonded to it.

All complex ions are Lewis adducts; the metal ions act as Lewis acids (electron-pair acceptors) and the ligands are Lewis bases (electron-pair donors). Only species with at least one lone-pair electrons can be a ligand. The Formation of Complex Ion

In aqueous solutions, metal ions form complex ions with water molecules as ligands. When another ligand is introduced into the solution, ligand exchanges occur and equilibrium is established. For example, when NH3 is added to aqueous solution containing Cu2+ ion, the following equilibrium occurs:

Cu(H2O)62+

(aq) + 4 NH3(aq) ⇌ [Cu(NH3)4]2+

(aq) + 6H2O

Kf = 4

3262

243

]][NHO)[Cu(H

])[Cu(NH

At molecular level, the ligand exchange process occurs in stepwise manner; water molecule is replace with NH3 molecule one at a time to give a series of intermediate species, each with its own formation constant. For convenience, the water molecules can be omitted from the equation.

1. Cu2+(aq) + NH3(aq) ⇌ Cu(NH3)

2+(aq);

Kf1 = ]][NH[Cu

])[Cu(NH

32

23

2. Cu(NH3)2+

(aq) + NH3(aq) ⇌ Cu(NH3)22+

(aq);

Kf2 = ]][NH)[Cu(NH

])[Cu(NH

32

3

223

3. Cu(NH3)32+

(aq) + NH3(aq) ⇌ Cu(NH3)42+

(aq);

Kf4 = ]][NH)[Cu(NH

])[Cu(NH

3233

243

The overall formation constant is the product of all intermediate formation constants:

Kf = Kf1 x Kf2 x Kf3 x Kf4 = 4

32

243

]][NH[Cu

])[Cu(NH

Kf is called the formation constant for the complex equilibrium. Sample problem-4: If a 30.0-mL solution containing 0.020 M Cu2+ is mixed with 20.0 mL of 0.20 M NH3

solution, what is the concentration of Cu2+ in solution? Kf = 5.0 x 1012



28

Solution: First calculate the concentration of each species in the solution after mixing, but before formation of complex ion Cu(NH3)4

2+:

[Cu2+] = 0.020 M x (30.0 mL/50.0 mL) = 0.012 M

[NH3] = 0.20 M x (20.0 mL/50.0 mL) = 0.080 M

Write the equilibrium expression:

Cu2+(aq) + 4 NH3(aq) ⇌ Cu(NH3)4

2+(aq) + 4 H2O

Since Kf is very large, we can assume that all of Cu2+ is converted to Cu(NH3)42+.

Then, [Cu(NH3)42+] = 0.012 M and [NH3] = 0.080 M � (4 x 0.012 M) = 0.032 M

Next, consider the following (reverse) equilibrium:

Cu(NH3)42+

(aq) ⇌ Cu2+(aq) + 4NH3(aq)

Kc = ])[Cu(NH

]][NH[Cu243

43

2

= 1/Kf = 1/(5.0 x 1012) = 2.0 x 10-13

Set up the following equilibrium table:

Concentration (M) Cu(NH3)42+

(aq) ⇌ Cu2+(aq) + 4 NH3(aq)

Initial: 0.012 M 0.000 0.032 M Change: - x + x + x Equilibrium: (0.012 � x) x (0.032 + x)

Since Kc is very small, we assume that x << 0.012 M,

Then, (0.012 � x) ~ 0.012, and (0.032 + x) ~ 0.032

Kc = (0.012)

0.032))(( 4x = 2.0 x 10-13;

x = [Cu2+] = 4

-13

(0.032)

)(0.012)10 x 0.2( = 2.3 x 10-9 M

Formation of Complex Ions and the Effect on Solubility

Ligand increases the solubility of slightly soluble ionic compounds if they form complex ions with the metal ions. For example, silver chloride, AgCl, is more soluble in ammonia solution because silver ions form complex ions with NH3:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10;

Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2

+(aq); Kf = 1.7 x 107;

AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+

(aq) + Cl-(aq);

Knet = Ksp x Kf = (1.6 x 10-10)(1.7 x 107) = 2.7 x 10-3 Exercise: 1. Calculate the solubility (a) in mol/L and (b) in g/L of AgCl in 2.0 M NH3(aq). What is the

concentration of Ag+ in this solution? (Ksp = 1.6 x 10-10; Kf = 1.7 x 107 for Ag(NH3)2+)

2. Will AgCl precipitates form if 5.0 mL of 0.10 M AgNO3 is added to 5.0 mL of solution

containing 0.10 NaCl and 2.0 M NH3? (b) Will precipitate form if [NH3] = 1.0 M?



29

3. Calculate the concentrations of Ag+, Ag(S2O3)-, and Ag(S2O3)2

3- in a solution prepared by mixing 20.0 mL of 0.010 M AgNO3 and 30.0 mL of 2.0 M Na2S2O3 ?

Ag+(aq) + S2O3

2-(aq) ⇌ Ag(S2O3)

-; Kf1 = 7.4 x 108

Ag(S2O3)-(aq) + S2O3

2-(aq) ⇌ Ag(S2O3)2

3-(aq); Kf2 = 3.9 x 104

Ag+(aq) + 2S2O3

2-(aq) ⇌ Ag(S2O3)2

3-(aq); Kf = 2.9 x 1013

Miscellenius Problems PROBLEM. 1. A chemist needs to prepare about 400 mL of a buffer solution of pH 5.0 using acetic acid. What should the weight of sodium acetate (MW 82.034 g/mole) should be added to 400 mL of 0.1 M acetic acid (MW 60.052 g/mole) solution to prepare the buffer? (Ka=1.75X10-5 for acetic acid.)

Ans 5.76 gram

PROBLEM. 2 20 mL of a diprotic acid, H2A, is titrated with 0.1 M NaOH titrant. The resulting titration curve is shown below.

(a) Label the two points where the solution has the maximum buffer capacity and label first and second equivalence points as .

(b) Estimate the formal concentration of H2A from the graph.

(c) There are simple expressions for the pH at points A, B, and C. Give the equations below.

(d) Estimate Ka1 and Ka2 for this acid from your estimate of pH at points A, B, and C.

(e) At what indicator pH would you want your color-change indicator to have in order to titrate to the second equivalence point?

Ans (a) Buffer capacity at A & C

1st equivalence point at B 2nd at D.

(b) 0.1 M

©



30

(D)

(e)At pH=10

PROBLEM. 3.Calomel, Hg2Cl2, (MW 472.09) dissolves to form Hg22+ and chloride with

a Ksp=1.2 10-18. Calculate the mass of calomel that will dissolve in...

A. 1.00 L pure water Ans. 3.16 x 10-4 gram

The equilibrium equation is Hg2Cl2 s)<=>Hg22+ (aq) + 2Cl- (aq)

B. 1.00 L of a 0.030 M solution of NaCl . ans 6.29 x 10 -13 gram

PROBLEM. 4. Which of the following would be most suitable for preparing a buffer of pH 2.5? Show your work and briefly state why you chose that species.

1) H2C2O4 (Ka1=5.60 10-2 Ka2=5.42 10-5) 2) H3PO4 (Ka1=7.11 10-3 Ka2=6.32 10-8)

3) B(OH)3 (Ka1=5.81 10-10 Ka2=1.82 10-13) 4) H2CO3 (Ka1=4.45 10-7 Ka2=1.69 10-9)

Ans

The best acid to use is the one with a pKa= pH. Of the choices, H3PO4, with pKa1=2.15, is the closest to the target value of pH=2.5

Problem-5 A solution was prepared by addition of 0.050 mole NaOH to 500 mL of a 0.10 M acetic acid (Ka=1.75 10-5) solution and then diluted to 1 L. Find the pH of the resulting buffer. Ans 8.7

Problem-6 . 50 mL of a solution containing Cl- and Br- are titrated with 0.01 M AgNO3 solution. The two equivalence points, found using an electrochemical apparatus, occur at 12.5 mL and 18.0 mL total volume of titrant added. What are the molar concentrations of Cl- and Br- in the unknown solution? The silver-halide solubility products are 1.8 10-10 for AgCl and 5.0 10-13 for AgBr.

Ans Br- =0.0025 M Cl-= 0.0011 M

PROBLEM-6 . Lanthanum iodate, La(IO3)3, (M.W. 663.62) has Ksp=1.010-11. How many grams lanthanum iodate will dissolve in...

A) 1.00 L pure water ?

The equilibrium equation is La(IO3)3 (s) La3+ (aq) + 3IO3- (aq) Ans 0.518 gram

B) A solution is prepared by dissolving 9.88 g trichloroacetic acid (M.W. 163.39) in 500 mL solution. At this concentration, the acid is 70% dissociated. What is the Molarity of trichloroacetate ion? Ans 0.036 Molar

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Ionic Equillibrium