Applications of Aqueous Applications of Aqueous EquilibriaEquilibria

Chapter 8Chapter 8

Chapter 8: Applications of Aqueous Equilibria

8.1 Solutions of Acids or Bases Containing a Common Ion8.2 Buffered Solutions8.3 Exact Treatment of Buffered Solutions8.4 Buffer Capacity8.5 Titrations and pH Curves8.6 Acid-Base Indicators8.7 Titration of Polyprotic Acids8.8 Solubility Equilibria and The Solubility Product8.9 Precipitation and Qualitative Analysis8.10 Complex Ion Equilibria

A base swirling in a solution containing phenolphthalein

Like Example 8.1 (P 274-5) - INitrous acid, a very weak acid, is only 2.0% ionized in a 0.12 M solution. Calculate the [H+], the pH, and the percent dissociation of HNO2 in a 1.0 M solution that is also 1.0 M in NaNO2!

HNO2(aq) H+(aq) + NO2

-(aq)

Ka = = 4.0 x 10-4[H+] [NO2 -] [HNO2]

Initial Concentration (mol/L) Equilibrium Concentration (mol/L)

[HNO2]0 = 1.0 M [HNO2] = 1.0 – x(from dissolved HNO2)[NO2

-]0 = 1.0 M [NO2-] = 1.0 + x

(from dissolved NaNO2)[H+]0 = 0 [H+] = x(neglect the contribution from water)

Like Example 8.1 (P 274-5) - II

Ka = = = 4.0 x 10-4[H+] [NO2

-] [HNO2]

( x ) ( 1.0 + x ) (1.0 – x )

Assume x is small as compared to 1.0:

X (1.0) (1.0) = 4.0 x 10-4 or x = 4.0 x 10-4 = [H+]

Therefore pH = - log [H+] = - log ( 4.0 x 10-4 ) = 3.40

The percent dissociation is:

4.0 x 10-4

1.0 x 100 = ________ %

Nitrous acid Nitrous acid alone + NaNO2

[H+] 2.0 x 10-2 4.0 x 10-4

pH 1.70 3.40% Diss 2.0 0.040

Example 8.2 (P276-9) - IA buffered solution contains 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and o.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution, and the pH when 0.010 M of solid NaOH is addedto 1.0 L of this buffer and to pure water.

HC2H3O2 (aq) H+(aq) + C2H3O2 (aq)

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

-] [HC2H3O2]

Initial Concentration (mol/L) Equilibrium Concentration (mol/L)

[HC2H3O2]0 = 0.50 [HC2H3O2] = 0.50 – x[C2H3O2

-]0 = 0.50 [C2H3O2-] = 0.50 +x

[H+]0 = 0 [H+] = x

X mol/L of HC2H3O2

dissociates to reach equilibrium

~

Example 8.2 (P276-9) - II

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

-] [HC2H3O2]

( x ) ( 0.50 + x) 0.50 - x

(x) (0.50) 0.50

x = 1.8 x 10-5 The approximation by the 5% rule is fine:

[H+] = x = 1.8 x 10-5 M and pH = 4.74

To calculate the pH and concentrations after adding the base:

OH- + HC2H3O2 H2O + C2H3O2-

Beforereaction: 0.010 mol 0.50 mol - 0.50 mol

After reaction: 0.010 – 0.010 0.50 – 0.10 - 0.50 + 0.10 = 0 mol = 0.49 mol = 0.51 mol

Note that 0.01 mol of acetic acid has been converted to acetate ionby the addition of the base.

~

Example 8.2 (P276-9) - IIIInitial Concentration (mol/L) Equilibrium concentration (mol/L)

[HC2H3O2]0 = 0.49 [HC2H3O2] = 0.49 – x[C2H3O2

-]0 = 0.51 [C2H3O2-] = 0.51 + x

[H+]0 = 0 [H+] = x

X mol/L of HC2H3O2

Dissociates to reach equilibrium

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

-]

[HC2H3O2]

(x)(0.51+ x) 0.49 - x

(x)(0.51) 0.49

x = 1.7 x 10-5 and pH = 4.76

If the base is added to pure water without the buffer being presentwe get an entirely different solution:If the 0.01 mol of NaOH is added to 1.0 L of pure water theConcentration of hydroxide ion is 0.01 M.

[H+] = = = __________ and the pH = ______ Kw

[OH-]1.0 x 10-14

1.0 x 10-2

ModifiedpH

Step 1:Do stoichiometric calculations todetermine new concentrations.assume reaction with H+/OH-

goes to completion.

Originalbufferedsolution

pH

Step 2:Do equilibrium

calculations.

(H+/OH- added)

Le Châtelier’s principle for the dissociation equilibrium for HF

HF(aq) H+(aq) + F-

(aq)

Molecular model: F-, Na+, HF, H2O

How Does a Buffer Work

Lets add a strong base to a weak acid and see what happens:

OH- + HA A- + H2O

Original buffer pH

Final pH of bufferclose to original-

Added OH- ionsReplaced by A- ions

Ka =[H+] [A-] [HA] [H+] = Ka

[HA] [A-]

The Effect of Added Acetate Ion on the Dissociation of Acetic Acid

[CH3COOH] [CH3COO-]added % Dissociation* pH

0.10 0.00 1.3 2.89

0.10 0.050 0.036 4.44

0.10 0.10 0.018 4.74

0.10 0.15 0.012 4.92

* % Dissociation = x 100 [CH3COOH]dissoc

[CH3COOH]init

Human blood is a buffered solution

Source: Visuals Unlimited

How a Buffer Works–IA buffer consists of a solution that contains “high” concentrations of the acidic and basic components. This is normally a weak acid and the anion of that weak acid, or a weak base and the corresponding cation of the weak base. When small quantities of H3O+ or OH- are added to the buffer, they cause a small amount of one buffer component to convert into the other. As long as the amounts of H3O+ and OH- are small ascompared to the concentrations of the acid and base in the buffer, the added ions will have little effect on the pH since they are consumed by the buffer components.Consider a buffer made from acetic acid and sodium acetate:

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+

(aq)

Ka = or [H3O+] = Ka x[CH3COO-] [H3O+]

[CH3COOH]

[CH3COOH]

[CH3COO-]

How a Buffer Works–IILet’s consider a buffer made by placing 0.25 mol of acetic acid and 0.25 mol of sodium acetate per liter of solution. What is the pH of the buffer? And what will be the pH of 100.00 mL of the buffer before and after 1.00 mL of concentrated HCl (12.0 M) is added to the buffer? Whatwill be the pH of 300.00 mL of pure water if the same acid is added?

[H3O+] = Ka x = 1.8 x 10-5 x = 1.8 x 10-5 [CH3COOH]

[CH3COO-] (0.25)

(0.25)

pH = -log[H3O+] = -log(1.8 x 10-5) = pH = ____ Before acid added!

1.00 mL conc. HCl 1.00 mL x 12.0 mol/L = 0.012 mol H3O+

Added to 300.00 mL of water :

0.012 mol H3O+

301.00 mL soln.= 0.0399 M H3O+

pH = -log(0.0399 M) pH = _____ Without buffer!

How a Buffer Works–IIIAfter acid is added: Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+

Initial 0.250 ---- 0.250 0Change +0.012 ---- -0.012 0.012Equilibrium 0.262 ---- 0.238 0.012

Solving for the quantity ionized:

Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+

Initial 0.262 ---- 0.238 0Change -x ---- +x +xEquilibrium 0.262 - x ---- 0.238 + x x

[CH3COOH]

[CH3COO-][H3O+] = Ka x =1.8 x 10-5 x = 1.982 x 10-5(0.262)

(0.238)

Assuming: 0.262 - x = 0.262 & 0.238 + x = 0.238

pH = -log(1.982 x 10-5) = 5.000 - 0.297 = ____ After the acid is added!

How a Buffer Works–IV

Suppose we add 1.0 mL of a concentrated base instead of an acid. Add1.0 mL of 12.0 M NaOH to pure water and our buffer, and let’s see what the impact is: 1.00 mL x 12.0 mol OH-/1000mL = 0.012 mol OH-

This will reduce the quantity of acid present and force the equilibrium to produce more hydronium ion to replace that neutralized by the addition of the base!

Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+

Initial 0.250 ---- 0.250 0Change - 0.012 ---- +0.012 +0.012Equilibrium 0.238 ---- 0.262 +0.012

Assuming: Again, using x as the quantity of acid dissociated we get: our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238

[H3O+] = 1.8 x 10-5 x = 1.635 x 10-50.2380.262

pH = -log(1.635 x 10-5) = 5.000 - 0.214 = ____ After base is added!

How a Buffer Works–V

By adding the 1.00mL base to 300.00 mL of pure water we would get a hydroxide ion concentration of:

301.00 mL0.012 mol OH-

[OH-] = = 3.99 x 10-5 M OH-

This calculates out to give a pH of:

The hydrogen ion concentration is:

[H3O+] = = = 2.506 x 10-10Kw

[OH-]1 x 10-14

3.99 x 10-5 M

pH = -log(2.5 6 x 10-10) = 10.000 - 0.408 = 9.59 With 1.0 mL of the base in pure water!

In summary: Buffer alone pH = 4.74 Buffer plus 1.0 mL base pH = 4.79 Base alone pH = 9.59 Buffer plus 1.0 mL acid pH = 4.70 Acid alone pH = 1.40

Molecular model: HC2H3O2, C2H3O2-

The Relation Between Buffer Capacity and pH Change

A digital pH meter shows the pH of the buffered solution to be 4.74

Preparing a BufferProblem: The ammonia-ammonium ion buffer has a pH of about 9.2 and can be used to keep solutions in the basic pH range. What mass of ammonium chloride must be added to 400.00 mL of a 3.00 M ammonia solution to prepare a buffer ?Plan: The conjugate pair is the ammonia-ammonium ion pair which has an equilibrium constant Kb = 1.8 x 10 -5. The reaction equation with water can be written along with the Kb expression, since we want to add sufficient ammonium ion to equal the aqueous ammonia concentration.Solution: The reaction for the ammonia-ammonium ion buffer is:

NH3 (aq) + H2O(l) NH4+

(aq) + OH-(aq)

Kb = = 1.8 x 10-5[NH4

+] [OH-]

[NH3]

[NH4+] = 3.00 mol x 0.400 L = 1.20 mol

NH4Cl = 53.49 g/mol

Therefore mass =NH4Cl= 1.20 mol x 53.49g/molmass = _________ g NH4Cl

L

Pure water at pH 7.00

The Henderson-Hasselbalch Equation

Take the equilibrium ionization of a weak acid:

HA(aq) + H2O(aq) = H3O+(aq) + A-

(aq) Ka =[H3O+] [A-]

[HA]Solving for the hydronium ion concentration gives:

[H3O+] = Ka x[HA][A-]

Taking the negative logarithm of both sides:

-log[H3O +] = -log Ka - log ( )[HA] [A-]

pH = -log Ka - log( )[HA] [A-]

Generalizing for any conjugate acid-base pair :

pH = log Ka + log( )[base][acid]

Henderson-Hasselbalch equation

When 0.01 mol NaOH is added to 1.0 L of pure water, the pH jumps to 12.00

Like Example 8.3 (P 282-3) -IProblem: Instructions for making a buffer say to mix 60.0 ml of0.100 M NH3 with 40.0 ml of 0.100 M NH4Cl. What is the pH of this buffer?

The combined volume is 60.0 ml + 40.0 ml = 100.0 ml

Moles of Ammonia = VolNH3 x MNH3 = 0.060 L x 0.100 M = 0.0060 mol

Moles of Ammonium ion = VolNH4Cl x MNH4Cl = 0.040 L x 0.100 M = = 0.0040 mol

[NH3] = = 0.060 M ; [NH4+] = = 0.040 M

0.0060 mol 0.100 L

0.0040 mol 0.100 L

Concentration (M) NH3 (aq) + H2O(l) NH4+

(aq) + OH-(aq)

Starting 0.060 0.040 0Change -x +x +xEquilibrium 0.060 – x 0.040 – x x

Like Example 8.3 (P 282-3) - II

Substituting into the equation for Kb:

Kb = = 1.8 x 10-5 = [NH4

+] [OH-] [NH3]

(0.040 + x) (x) (0.060 – x)

Assume : 0.060 – x = 0.060 ; 0.040 + x = 0.040~~

Kb = 1.8 x 10-5 = x = 2.7 x 10-5 0.040 (x) 0.060

Check assumptions: 0.040 + 0.000027 = 0.040 or 0.068% 0.060 – 0.000027 = 0.060 or 0.045%

[OH-] = 2.7 x 10-5 ; pOH = - log[OH-] = - log (2.7 x 10-5) = 5 – 0.43 pOH = 4.57

pH = 14.00 – pOH = 14.00 – 4.57 = __________

pH Box

pH [H3O+]

pOH [OH-]pOH = -log[OH-]

[OH-] = 10-pOH

pH = -log[H3O+]

[H3O+] = 10-pH

Kw = 1 x 10-14

@ 25oC[H3O+][OH-]= =1 x 10-14

pH + pOH == 14 @ 25oC

When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first.

After the stoichiometric calculations are completed, then consider the equilibrium

calculations. This procedure can be represented as follows:

Summary: Characteristics of Buffered Solutions

Buffered solutions contain relatively large concentrations of a weak acid and its corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+.

When H+ is added to a buffered solution, it reacts essentially to completion with the weak base present: H+ + A- HA or H+ + B BH+

When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present. OH- + HA A- + H2O or OH- + BH+ B + H2O

The pH of the buffered solution is determined by the ratio of the concentrations of the weak base and weak acid. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A- or G and BH+) are largecompared with the amounts of H+ or OH- added.

OH- ions are not allowed to accumulate but are replaced by A- ions.

Exact Treatment of Buffer SolutionsWe can use several relationships to calculate the exact solution to buffered solution problems:

Charge – balance equation: [Na+] + [H+] = [A-] + [OH-]Material – Balance equation: [A-]0 + [HA]0 = [HA] + [A-]

Since [A-]0 = [Na+] and Kw = [OH-][H+] , we can rewrite the chargebalance equation, and solve for [A-] :

[A-] = [A-]0 +[H+]2 – Kw

[H+]From the mass balance equation solved for [HA] we get:

[HA] = [A-]0 + [HA]0 – [A-]Substituting the expression for [A-], and substituting into the Ka

expression for HA we obtain:

Ka = =[H+][A-] [HA]

[H+]{ [A-]0 + } [H+]2 – Kw

[H+]

[H+]2 – Kw

[H+][HA]0 -

When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the [HA]/[A-] ratio and thus the

[H+] stay virtually constant.

Example 8.4 (P 286-7) - I

Calculate the pH of a buffered solution containing 3.0 x 10-4 M HOCl(Ka = 3.5 x 10-8) and 1.0 x 10-4 M NaOCl.

Ka = = 3.5 x 10-8[H+] [OCl-] [HOCl]

Let x = [H+] then:[OCl-] = 1.0 x 10-4 + x[HOCl] = 3.0 x 10-4 - x

3.5 x 10-8 = =[H+] [OCl-] [HOCl]

(x)(1.0 x 10-4 + x) (3.0 x 10-4 – x )

Assuming x is small compared to 1.0 x 10-4 and solving for x we have:

[H+] = x = = 1.05 x 10-7 M = 1.1 x 10-7 M1.05 x 10-11

1.0 x 10-4

Since this is close to that of water we must use the equation that uses water, and takes it’s ionization into account.

Example 8.4 (P 286-7) - II

Ka = 3.5 x 10-8 =

[H+]{[OCl-]0 + }[H+]2 – 1.0 x 10-14

[H+]

[H+]2 – 1.0 x 10-14

[HOCl]0 -[H+]

Where:[OCl-]0 = 1.0 x 10-4 M[HOCl]0 = 3.0 x 10-4 M

We expect [H+] to be close to 1.0 x 10-7, so [H+]2 to be about 1.0 x 10-14

[H+]2 – 1.0 x 10-14[OCl-]0 = 1.0 x 10-4 M >>>

[H+]

[H+]2 – 1.0 x 10-14[HOCl]0 = 1.0 x 10-4 M >>>

[H+]The expression becomes:[H+][OCl-] [HOCl]

3.5 x 10-8 = =[H+](1.0 x 10-4) (3.0 x 10-4)

[H+][OCl-] [HOCl]

3.5 x 10-8 = =[H+](1.0 x 10-4) (3.0 x 10-4)

Example 8.4 (P 286-7) - III

[H+] = 1.05 x 10-7 M = 1.1 x 10-7 M

Using this result, we can check the magnitude of the neglected term:

[H+]2 – 1.0 x 10-14

[H+]=

(1.05 x 10-7)2 – 1.0 x 10-14

1.05 x 10-7=__________

This result suggests that the approximation was fine!

Molecular model: Cl-, NH4+

pH and Capacity of Buffered Solutions

The pH of a buffered solution is determined by the ratio[A-]/[HA].

The capacity of a buffered solution is determined by the magnitudesOf [HA] and [A-]

Original solution and new solution

Original solution and new solution

Solution A and Solution B

Example 8.5 (P 287-9) - ICalculate the change in pH that occurs when 0.010 mol of gaseousHCl is added to 1.0 L of each of the following solutions: Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2

Solution B: 0.050 M HC2H3O2 and 0.0500 M NaC2H3O2

For Acetic acid, Ka = 1.8 x 10-5

Use the Henderson-Hasselbalch equation for initial pH:

pH = pKa + log{ } [C2H3O2

-][H C2H3O2]

Since [C2H3O2-] = [H C2H3O2]

The equation becomes:

pH = pKa + log (1) = pKa = -log(1.8 x 10-5) = 4.74

Adding 0.010 mol of HCl will cause a shift in the equilibrium due to:

H+(aq) + C2H3O2

-(aq) H C2H3O2 (aq)

Example 8.5 (P 287-9) - II

For Solution A: H+ + C2H3O2- H C2H3O2

Before reaction 0.010 M 5.00 M 5.00 MAfter reaction 0 4.99 M 5.01 M

Calculate the new pH using the Henderson-Hasselbalch equation:

pH = pKa + log ( ) = 4.74 + log ( ) = 4.74 – 0.0017 = _____

[C2H3O2-]

[H C2H3O2]

4.995.01

For Solution B: H+ + C2H3O2- H C2H3O2

Before reaction 0.010 M 0.050 M 0.050 MAfter reaction 0 0.040 M 0.060 M

The new pH is: pH = 4.74 + log( ) = 4.74 – 0.18 = ______0.0400.060

A burette valve

Source: American Color

Figure 8.1: The pH curve for the titration of 50.0

Vol NaOH added (mL)

Figure 8.2: The pH curve for the titration of 100.0

Treat the stoichiometry

Summary: Titration Curve Calculations

A Stoichiometry problem. The reaction of hydroxide ion withthe weak acid is assumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined.

An equilibrium problem. The position of the weak acid equilibrium is determined, and the pH is calculated.

Figure 8.3: The pH curve for the titration of 50.0

Weak acid

Calculating the pH During a Weak Acid-Strong Base Titration–I

Problem: Calculate the pH during the titration of 20.00 mL of 0.250 Mnitrous acid (HNO2; Ka = 4.5 x 10-4) after adding different volumes of0.150 M NaOH : (a) 0.00 mL (b) 15.00 mL (c) 20.00 mL (d) 35.00 mL.Plan: (a) We just calculate the pH of a weak acid. (b)-(d) We calculate the amounts of acid remaining after the reaction with the base, and the anion concentration, and plug these into the Henderson-Hasselbalch eq.Solution: HNO2 (aq) + NaOH(aq) H2O(l) + NaNO2 (aq)

HNO2 (aq) + H2O(l) H3O+(aq) + NO2

-(aq)

(a) Ka = = = 4.5 x 10-4

[H3O+] [NO2-]

[HNO2]x (x)

0.250 Mx2 = 1.125 x 10-4

x = 1.061 x 10-2

pH = -log(1.061 x 10-2) = 2.000 - 0.0257 = ________ no base added

pH = pKa + log = 3.35 + log(0.00225/0.00275) ( ) [NO2-]

[HNO2]

Calculating the pH During a Weak Acid-Strong Base Titration–II

(b) 15.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which willneutralize (15.00 mL x 0.150 mmol/mL = 2.25 mmol of HNO2), leaving 2.75 mmol HNO2, and generating 2.25 mmol of nitrite anion.

HNO2 (aq) + H2O(l) H3O+(aq) + NO2

-(aq)Concentration (M)

Initial 0.00275 ---- 0 0.00225Change -x ---- +x +xEquilibrium 0.00275 - x ---- x 0.00225 + x

pH = 3.35 -0.0872 = _________ with 15.0 mL of NaOH added

Calculating the pH During a Weak Acid-Strong Base Titration–III(c) 20.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which willneutralize (20.00 mL x 0.150 mmol/mL = 3.00 mmol of HNO2), leaving 2.00 mmol HNO2, and generating 3.00 mmol of nitrite anion.

HNO2 (aq) + H2O(l) H3O+(aq) + NO2

-(aq)Concentration (M)

Initial 0.00200 ---- x 0.00300

pH = pKa + log = 3.35 + log(0.00300/0.00200) ( ) [NO2-]

[HNO2]

pH = 3.35 + 0.176 = __________ with 20.00 mL of base added

(d) 35.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which willneutralize (35.00 mL x 0.150 mmol/mL = 5.25 mmol of HNO2), leaving no HNO2, and generating 5.00 mmol of nitrite anion. There will be an excess of 0.25 mmol of NaOH which will control the pH.

Calculating the pH During a Weak Acid-Strong Base Titration–IV

(d) continued Since all of the HNO2 has been neutralized, we only have to look at the concentration of hydroxide ion in the total volume of the solution to calculate the pH of the resultant solution.

combined volume = 20.00 mL + 35.00 mL = 55.00 mL

[OH-] = =0.004545 M 0.000250 mol OH-

0.05500 L

[H3O+] = = = 2.200 x 10-12Kw

[OH-]1 x 10-14

0.004545

pH = -log (2.200 x 10-12) = 12.000 - 0.342 = 11.66 when all of the acid neutralized, and there is an excess of NaOH

Figure 8.4: The pH curves for the titrations of 50.0

Figure 8.5: The pH curve for the titration of 100.0

Figure 8.6: The indicator phenolphthalein is pink

in basic solution and colorless in acidic solution.

Figure 8.7: (a) Yellow acid form of bromthymol blue; (b) a greenish tint is seen when the

solution contains 1 part blue and 10 parts yellow; (c) blue basic form.

Figure 8.8: The useful pH ranges for several common indicators

Colors and Approximate pH Range of Some Common Acid-Base Indicators

Figure 8.9: pH of .10 M HCI

Figure 8.10: pH/.1 M HC2H3O2

Figure 8.11: A summary of the important equilibria at various points in the titration of a

triprotic acid

An X ray of the upper gastrointestinal

Source: Photo Researchers, Inc.

Equilibria of Slightly Soluble Ionic Compounds

When a solution becomes saturated and a precipitate forms we move into the area of insoluble material in solution, and we begin to calculate thequantity of material that remains in solution. We are working with what we call the: “Solubility Product”

The equilibrium constant that is used for these calculations is called the Solubility-product constant: Ksp

Example : Lead chromate PbCrO4 (s) Pb2+(aq) + CrO4

2-(aq)

Qc =[Pb2+][CrO4

2-]

[PbCrO4]

Since the concentration of a solid is constant,we can move it to the other side of the equalssign and combine it with the constant yieldingthe solubility product constant Ksp[PbCrO4] x Qc = [Pb2+][CrO4

2-] = Ksp

Precipitation of bismuth

Like Example 8.12 (P 319)The Ksp value for the mineral fluorite, CaF2 is 3.4 x 10-11 . CalculateThe solubility of fluorite in units of grams per liter.

Concentration (M) CaF2 (s) Ca2+(aq) + 2 F-

(aq)

Starting 0 0Change +x +2xEquilibrium x 2x

Substituting into Ksp:[Ca2+][F-]2 = Ksp

(x) (2x)2 = 3.4 x 10-11

4x3 = 3.4 x 10-11

x =

x = 2.0 x 10-4

3.4 x 10-11

43

The solubility is 2.0 x 10-4 moles CaF2 per liter of water. To get masswe must multiply by the molar mass of CaF2 (78.1 g/mol).

2.0 x 10-4 mol CaF2 x = _________ g CaF2 per L78.1 g CaF2

1 mol CaF2

Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds

Problem: Write the ion-product expression for (a) silver bromide;(b) strontium phosphate; (c) aluminum carbonate; (d) nickel(III) sulfide.Plan: Write the equation for a saturated solution, then write the expression for the solubility product.Solution:

(a) Silver bromide: AgBr(s) Ag+(aq) + Br -

(aq)

Ksp = [Ag+] [Br -]

(b) Strontium phosphate: Sr3(PO4)(s) 3 Sr2+(aq) + 2 PO4

3-(aq)

Ksp = [Sr2+]3[PO43-]2

(c) Aluminum carbonate: Al2(CO3)3 (s) 2 Al3+(aq) + 3 CO3

2-(aq)

Ksp = [Al3+]2[CO32-]3

(d) Nickel(III) sulfide: Ni2S3 (s) + 3 H2O(l) 2 Ni3+

(aq) + 3 HS -(aq) + 3 OH-

(aq)

Ksp =[Ni3+]2[HS-]3[OH-]3

Table 8.5: Ksp Values at 25 C for Common Ionic Solids

Determining Ksp from Solubility

Problem: Lead chromate is an insoluble compound that at one time was used as the pigment in the yellow stripes on highways. It’s solubility is4.33 x 10 -6g/100mL water. What is the Ksp?Plan: We write an equation for the dissolution of the compound to see the number of ions formed, then write the ion-product expression.Solution:

PbCrO4 (s) Pb2+(aq) + CrO4

2-(aq)

Molar solubility of PbCrO4 = x x 4.33 x 10 -6g100 mL

1000 ml 1 L

1mol PbCrO4

323.2 g

= 1.34 x 10 -8 M PbCrO4

1 Mole PbCrO4 = 1 mole Pb2+ and 1 mole CrO42-

Therefore [Pb2+] = [CrO42-] = 1.34 x 10-8 M

Ksp = [Pb2+] [CrO42-] = (1.34 x 10 -8 M)2 = _______________

Determining Solubility from Ksp

Problem: Lead chromate used to be used as the pigment for the yellowlines on roads, and is a very insoluble compound. Calculate the solubility of PbCrO4 in water if the Ksp is equal to 2.00 x 10-16.Plan: We write the dissolution equation, and the ion-product expression.Solution: Writing the dissolution equation, and the ion-product expression: PbCrO4 (s) Pb2+

(aq) + CrO42-

(aq)

Ksp = 2.00 x 10-16 =[Pb2+][CrO42]

Concentration (M) PbCrO4 Pb2+ CrO42-

Initial ---------- 0 0Change ---------- +x +xEquilibrium ---------- x x

Ksp = [Pb2+] [CrO42-] = (x)(x ) = 2.00 x 10-16 x = 1.41 x 10-8

Therefore the solubility of PbCrO4 in water is 1.41 x 10-8 M

Relationship Between Ksp and Solubility at 25oC

No. of Ions Formula Cation:Anion Ksp Solubility (M)

2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4

2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4

2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5

3 Ca(OH)2 1:2 6.5 x 10-6 1.2 x 10-2

3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3

3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4

3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5

The Effect of a Common Ion on Solubility

PbCrO4(s) Pb2+(aq) + CrO4

2-(aq) PbCrO4(s) Pb2+

(aq) + CrO42-

(aq;

added)

Calculating the Effect of a Common Ion on Solubility

Problem: What is the solubility of silver chromate in 0.0600 M silvernitrate solution? Ksp = 2.6 x 10-12 .Plan: From the equation and the ion-product expression for Ag2CrO4, we predict that the addition of silver ion will decrease the solubility.Solution: Writing the equation and ion-product expression:

Ag2CrO4 (s) 2 Ag+(aq) + CrO4

2-(aq) Ksp = [Ag+]2[CrO4

2-]

Concentration (M) Ag2CrO4 (s) 2 Ag+(aq) + CrO4

2-(aq)

Initial --------- 0.0600 0Change --------- +2x +x Equilibrium --------- 0.0600 + 2x x

Assuming that Ksp is small, 0.0600 M + 2x = 0.600 M

Ksp = 2.6 x 10-12 = (0.0600)2(x) x = 7.22 x 10-10 M

Therefore, the solubility of silver chromate is 7.22 x 10-10 M

Test for the Presence of a Carbonate

Predicting the Effect on Solubility of Adding Strong Acid

Problem: Write balanced equations to explain whether addition of H3O+

from a strong acid affects the solubility of:(a) Iron(II) cyanide(b) Potassium bromide(c) Aluminum hydroxidePlan: Write the balanced dissolution equation and note the anion. Anions of weak acids react with H3O+ and shift the equilibrium position toward more dissolution. Strong acid anions do not react, so added acid has noeffect.Solution: (a) Fe(CN)2 (s) Fe2+

(aq) + 2 CN-(a) Increases solubility

We noted earlier that CN- ion reacts with water to form the weak acid HCN, so it would be removed from the solubility expression.(b) KBr(s) K+

(aq) + Br -(aq) No effect This occurs since Br- is the

anion of a strong acid, and K+ is the cation of a strong base.(c) Al(OH)3 (s) Al3+

(aq) + 3 OH-(aq) Increases solubility

The OH- is the anion of water, a very weak acid, so it reacts with the added acid to produce water in a simple acid-base reaction.

The Chemistry of Limestone Formation

Gaseous CO2 is in equilibrium with aqueous CO2 in natural waters:

CO2 (g) CO2 (aq)

H2O(l)

The concentration of CO2 is proportional to the partial pressure of CO2 (g) in contact with the water (Henry’s law; section 13.3):

[CO2 (aq)] (proportional to) PCO2

The reaction of CO2 with water produces H3O+:

CO2 (aq) + 2 H2O(l) H3O+(aq) + HCO3

-(aq)

Thus, the presence of CO2 (aq) forms H3O+, which increases the solubility of CaCO3:

CaCO3 (s) + CO2 (aq) + H2O(l) Ca2+(aq) + 2 HCO3

-(aq)

Predicting the Formation of a Precipitate: Qsp vs. Ksp

Qsp = Ksp : When a solution becomes saturated, no more solute will dissolve, and the solution is called “saturated.” There will be no changes that will occur.

Qsp > Ksp : Precipitates will form until the solution becomes saturated.

Qsp< Ksp : Solution is unsaturated, and no precipitate will form.

The solubility produce constant, Ksp, can be compared to the ion-productconstant, Qsp to understand the characteristics of a solution with respect to forming a precipitate.

Predicting Whether a Precipitate Will Form–I

Problem: Will a precipitate form when 0.100 L of a solution containing 0.055 M barium nitrate is added to 200.00 mL of a 0.100 Msolution of sodium chromate?Plan: We first see if the solutions will yield soluble ions, then we calculate the concentrations, adding the two volumes together to get the total volume of the solution, then we calculate the product constant (Qsp), and compare it to the solubility product constant to see ifa precipitate will form.Solution: Both Na2CrO4 and Ba(NO3) are soluble, so we will have Na+,CrO4

2-, Ba2+ and NO3- ions present in 0.300 L of solution. We change

partners, look up solubilities, and we find that BaCrO4 would be insoluble, so we calculate it’s ion-product constant and compare it to the solubility product constant of 2.1 x 10-10:

[Ba2+] = = ___________ M in Ba2+

For Ba2+: [0.100 L Ba(NO3)2] [0.55 M] = 0.055mol Ba2+ 0.055 mol Ba2+

0.300 L

Predicting Whether a Precipitate Will Form–II

Solution cont.

For CrO42- : [0.100 M Na2CrO4] [0.200 L] = 0.0200 mol CrO4

2-

[CrO42-] = = 0.667 M in CrO4

2-0.0200 mol CrO4

2-

0.300 liters

Qsp = [Ba2+] [CrO42-] =(0.183 M Ba2+)(0.667 M CrO4

2-) = 0.121

Since Ksp = 2.1 x 10-10 and Qsp = 0.121, Qsp >> Ksp and a precipitate will form.

Figure 8.12: Separation of Cu2+ and Hg2

+

Figure 8.13: Separation of common cations by precipitation

The Stepwise Exchange of NH3 for H2O in M(H2O)4

2+

Formation Constants (Kf) of Some Complex Ions at 25oC–I

Complex Ion Kf

Ag(CN)2- 3.0 x 1020

Ag(NH3)2+ 1.7 x 107

Ag(S2O3)23- 4.7 x 1013

AlF63- 4 x 1019

Al(OH)4- 3 x 1033

Be(OH)42- 4 x 1018

CdI42- 1 x 106

Co(OH)42- 5 x 109

Cr(OH)4- 8.0 x 1029

Cu(NH3)42+ 5.6 x 1011

Fe(CN)64- 3 x 1035

Fe(CN)63- 4.0 x 1043

Formation Constants (Kf) of Some Complex Ions at 25oC–II

Complex Ion Kf

Hg(CN)42- 9.3 x 1038

Ni(OH)42- 2 x 1028

Pb(OH)3 - 8 x 1013

Sn(OH)3 - 3 x 1025

Zn(CN)42- 4.2 x 1019

Zn(NH3)42+ 7.8 x 108

Zn(OH)42- 3 x 1015

Calculating the Concentrations of Complex Ions–I

Problem: A chemist converts Ag(H2O)2+ to the more stable form

Ag(NH3)2+ by mixing 50.0 L of 0.0020 M Ag(H2O)2

+ and 25.0 L of0.15 M NH3. What is the final [Ag(H2O)2

+]? Kf Ag(NH3)2+ = 1.7 x 107.

Plan: We write the equation and the Kf expression, set up the table for the calculation, then substitute into Kf and solve.Solution: Writing the equation and Kf expression:

Ag(H2O)2+

(aq) + 2 NH3 (aq) Ag(NH3)2+

(aq) + 2 H2O(l)

Kf = = 1.7 x 107[Ag(NH3)2

+]

[Ag(H2O)2+][NH3]2

Finding the initial concentrations:

[Ag(H2O)2+]init = = 1.3 x 10-3 M50.0 L (0.0020 M)

50.0 L + 25.0 L

[NH3]init = = __________ M25.0 L (0.15 M)50.0 L + 25.0 L

Calculating the Concentration of Complex Ions–II

We assume that all of the Ag(H2O)2+ is converted Ag(NH3)2

+, so we set up the table with x = [Ag(H2O)2

+] at equilibrium.Ammonia reacted = [NH3]reacted = 2(1.3 x 10-3 M) = 2.6 x 10-3 M

Concentration (M) Ag(H2O)2+

(aq) 2NH3 (aq) Ag(NH3)2+ 2 H2O(aq)

Initial 1.3 x 10-3 5.0 x 10-2 0 ----Change ~(-1.3 x 10-3) ~(-2.6 x 10-3) ~(+1.3 x 10-3) ----Equilibrium x 4.7 x 10-2 1.3 x 10-3 ----

Kf = = = 1.7 x 107 [Ag(H2O)2

+][NH3]2

[Ag(NH3)2+] 1.3 x 10-3

x(4.7 x 10-2)2

x = ______________ M = [Ag(H2O)2+]

The Amphoteric Behavior of Aluminum Hydroxide

Separating Ions by Selective Precipitation–IProblem: A solution consists of 0.10 M AgNO3 and 0.15 M Cu(NO3)2.Calculate the [I -] that would separate the metal ions as their iodides.Kspof AgI = 8.3 x 10-17; Kspof CuI = 1.0 x 10-12.Plan: Since the two iodides have the same formula type (1:1), we compare their Ksp values and we see that CuI is about 100,000 times moresoluble than AgI. Therefore, AgI precipitates first, and we solve for [I -]that will give a saturated solution of AgI.Solution: Writing chemical equations and ion-product expressions:

AgI(s) Ag+(aq) + I -

(aq) Ksp = [Ag+][I -]

CuI(s) Cu+(aq) + I -

(aq) Ksp = [Cu+][I -]H2O

H2O

Calculating the quantity of iodide needed to give a saturated solutionof CuI:

[I -] = = = ________________ MKsp

[Cu+]1.0 x 10-12

0.15 M

Separating Ions by Selective Precipitation–II

Thus, the concentration of iodide ion that will give a saturated solution of copper(I) iodide is 1.0 x 10-11 M, and this will not precipitate the copper(I) ion, but should remove most of the silver ion. Calculating the quantity of silver ion remaining in solution we get:

[Ag+] = = = 1.2 x 10-6 MKsp

[I -]8.3 x 10-17

6.7 x 10-11

Since the initial silver ion was 0.10 M, most of it has been removed,and essentially none of the copper(I) was removed, so the separation was quite complete. If the iodide was added as sodium iodide, you would have to add only a few nanograms of NaI to remove nearly all of the silver from solution:

6.7 x 10-11 mol I - x x = ____ ng NaI149.9 g NaI

mol NaI

1 molNaI

mol I -

The General Procedure for Separating Ions in Qualitative Analysis

Separation into Ion Groups

Ion Group 1: Insoluble chlorides Ag+, Hg2

2+, Pb2+

Ion Group 2: Acid-insoluble sulfides Cu2+, Cd2+, Hg2+, As3+, Sb3+, Bi3+, Sn2+, Sn4+, Pb2+

Ion Group 3: Base-insoluble sulfides and hydroxides Zn2+, Mn2+, Ni2+, Fe2+, Co2+ as sulfides, and Al3+, Cr3+ as hydroxides

Ion Group 4: Insoluble phosphates Mg2+, Ca2+, Ba2+

Ion Group 5: Alkali metal and ammonium ions Na+, K+, NH4

+

Tests to Determine the Presence of Cations in Ion Group 5

Na+ ions K+ ions OH - + NH4+ NH3 + H2O

plus litmus paper

A Qualitative

Analysis Scheme for

Ag+, Al3+, Cu2+ and

Fe3+

Figure 8.14: Separation of Group I ions

The Acid Rain Problem–I

Normally the pH of precipitation is controlled by the reaction of carbon dioxide with water to form carbonic acid, which keeps the pH of “clean” rain in the slightly acid range, of about 5.6:

CO2 (g) + H2O(l) H2CO3 (aq) H3O+(aq) + HCO3

-(aq)

Sulfurous acid is produced by the reaction of sulfur dioxide with water,and even though it is a weak acid, it does add to the acidity of rain. A greater worry is the oxidation of the SO2 to form SO3 which reacts with water to form the strong acid, sulfuric acid.

SO2 (g) + H2O(l) H2SO3 (aq)

2 SO2 (g) + O2 (g) 2 SO3 (g)

SO3 (g) + H2O(l) H2SO4 (aq)

The Acid Rain Problem–II

Another strong acid is formed from the nitrogen oxides that are produced in all internal combustion engines, that is nitric acid, which is not only a strong acid, but also a strongly oxidizing acid.

N2 (g) + O2 (g) 2 NO(g)

2 NO(g) + O2 (g) 2 NO2 (g)

3 NO2 (g) + H2O(l) 2 HNO3 (aq) + NO(g) (Final step in theOstwald process)

These acids together make the problem of acid rain so severe in the eastern United States, and far worse in other parts of the world. The average pH of rain in the eastern U.S. back in 1984 was 4.2. Sweden and Pennsylvania share second place with a pH of 2.7, but the record was in Wheeling, West Virginia where the pH was 1.8. Some areas of California also reach a pH of 1.6. The problems are global in nature.

Formation of Acidic Precipitation