Buffers/Titration Aqueous Equilibria - Icontent.njctl.org/courses/science/ap-chemistry/... · Aqueous Equilibria - I Slide 2 / 113 Review hydrolysis of salts Slide 3 / 113 1 The hydrolysis

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Buffers/Titration Aqueous Equilibria - I

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Review hydrolysis of salts

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1 The hydrolysis of a salt of a weak base and a strong acid should give a solution that is __________.

A weakly basic

B neutral

C strongly basic

D weakly acidic

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2 A solution of one of the following is acidic. The compound is _________.

A NaCl

B CH3COONa

C NH4Cl

D NH3

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3 Which salt would undergo hydrolyzes to form an acidic solution?

A KCl

B NaCl

C NH4Cl

D LiCl

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4 Which substance when dissolved in water will produce a solution with a pH greater than 7?

A CH3COOH

B NaCl

C NaC2H3O2

D HCl

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5 A basic solution would result from the hydrolysis of one of the ions in this compound. The compound is__________.

A NaNO3

B NH4Cl

C CH3COONa

D CaCl2

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6 A water solution of which compound will turn blue litmus red?

A K2CO3

B NH4Cl

C NaOH

D NaCl

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The Common-Ion Effect

· Consider an aqueous solution of acetic acid:

· If acetate ion is added to the solution, Le Châtelier 's Principle predicts that the equilibrium will shift to the left.

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

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“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”

In the following Sample Problem, compare the pH of a 0.20 M solution of HF with the pH of the same solution after some NaF is added to it.

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SAMPLE PROBLEM #1a) Calculate the pH of a 0.20 M HFsolution. The Ka for HF is 6.8 ´ 10−4.


HF(aq) H2O(l) H3O+(aq) F−(aq) Initial 0.20 M 0 0

Change -x +x +x

At Equilibrium 0.20-x x x

HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)

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6.8 x 10-4 = x2

(0.20)

Ka = [H3O+][F-][HF] = 6.8 x 10-4

x2 = (0.20) (6.8 ´ 10−4)

x = 0.012

So, [H+] = [F-] = 0.012

and pH = 1.93


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SAMPLE PROBLEM #1 (con't)b) Calculate the pH of a for the same 0.20 M HF solution that also contains 0.10 M NaF.

Ka for HF is 6.8 ´ 10−4.

Ka = [H3O+][F-][HF] = 6.8 x 10-4

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b) Calculate the pH of a for the same 0.20 M HF solution that also contains 0.10 M NaF.

· We use the same equation and the same Ka expression, and set up a similar ICE chart. · But because NaF is a soluble salt, it completely dissociates, so F- is not initially zero.

HF(aq) H2O(l) H3O+(aq) F−(aq) Initial 0.20 M 0 0.10 M

Change

at Equilibrium

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HF(aq) H2O(l) H3O+(aq) F−(aq) Initial 0.20 M 0 0.10 M

Change -x +x +x

At Equilibrium 0.20-x x 0.10 + x


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The Common-Ion EffectTherefore,

x = [H3O+] = 1.4 x 10−3

and

pH = −log (1.4 x 10−3)

pH = 2.87

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1.4 ´ 10−3 = x

6.8 x 10-4 = (0.10)(x)(0.20)

(0.20)(6.8 x 10-4)(0.10)

= x

Ka =

Remember what the "x" is that you are solving for!

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· Consider the two solutions we just examined in Sample Problem #1.

· Compare the following:

· How do these results support Le Chatelier's Principle?


Solution Final [H3O+] pH

HF 0.12

HF and NaF 1.4 x 10-3

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The Ka for acetic acid, CH3COOH or HC2H3O2, is 1.8 x 10-5.

SAMPLE PROBLEM #2

a) Calculate the pH of a 0.30 M acetic acid, CH3COOH.

b) Calculate the pH of a solution containing 0.30 M acetic acid and 0.10 M sodium acetate.


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The Ka for acetic acid, CH3COOH, is 1.8 x 10-5.

SAMPLE PROBLEM #2 - Answers

a) Calculate the pH of a 0.30 M acetic acid solution.


x2

1.8 x 10-5 =

Ka = [H3O+][C2H3O2-]

[HC2H3O2] = 1.8 x 10-5

(0.30)

x = [H3O+] = ______________

and pH = ___________

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The Ka for acetic acid, CH3COOH or HC2H3O2, is 1.8 x 10-5.

SAMPLE PROBLEM #2 - Answers (con't)b) Calculate the pH of a solution containing 0.30 M acetic acid and 0.10 M sodium acetate.We use the same Ka expression, except now the acetate ion concentration is not the same as [H3O+].


[H3O+] (0.10)1.8 x 10-5 =

Ka = [H3O+][C2H3O2-]

[HC2H3O2] = 1.8 x 10-5

(0.30)

So now, [H3O+] = ______________

and pH = ___________

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HC2H3O2 (aq) + H2O (l) ↔ C2H3O2 -

(aq) + H3O+(aq)

· Consider the two solutions we just examined in Sample Problem #2.

· Compare the following:

· How do these results support Le Chatelier's Principle?


Solution [OH-] pH

HC2H3O2

HC2H3O2 and NaC2H3O2

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SAMPLE PROBLEM #3Calculate the pH of the following solutions:

a) 0.85 M nitrous acid, HNO2b) 0.85 M HNO2 and 0.10 M potassium nitrite, KNO2

The Ka for nitrous acid is 4.5 x 10-4.


Answers: a) b)

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7 The ionization of HF will be decreased by the addition of:

A NaCl

B NaF

C HCl

D Both A and B

E Both B and C

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8 The dissociation of Al(OH)3 will be decreased by the addition of:

A KOH

B AlCl3C Mg(OH)2

D Both A and B

E A, B and C

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9 Which of the following substances will not decrease the ionization of H3PO4?

A K3PO4

B HCl

C Na3PO4

D None of them will decrease the ionization

E All of them will decrease the ionization

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The Common-ion effect can also be observed with weak bases.

Consider the ionization of ammonia, NH3, which is a weak base.

NH3 + H2O <---> NH4

+ + OH-

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Suppose a salt of the conjugate base is added to a solution of ammonia. What effect would this have on pH?

NH3 + H2O <---> NH4

+ + OH-

As with the previous Sample Problems, let us calculate the pH of the following:

a) a 0.45 M solution of NH3

b) a 0.45 M solution of NH3 that also contains 0.15 M NH4Cl, ammonium chloride -- a soluble salt that readily yields NH4

+ ions.

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a) a 0.45 M solution of NH3

b) a 0.45 M solution of NH3 that also contains 0.15 M NH4ClThe Kb for NH3 is 1.8 x 10-5.


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SAMPLE PROBLEM #4 - Answers

a) Calculate the pH of a 0.45 M solution of NH3. The Kb for NH3 is 1.8 x 10-5.

NH3 + H2O <---> NH4

+ + OH-


[NH4+][OH- ]

[NH3]Kb = 1.8 x 10-5 =

x = [OH-] = _____________

pH = _________________

x2

(0.45)1.8 x 10-5 =

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SAMPLE PROBLEM #4 - Answers (con't)

b) Calculate the pH of a 0.45 M solution of NH3 that also contains 0.15 M NH4Cl. The Kb for NH3 is 1.8 x 10-5.

NH3 + H2O <---> NH4

+ + OH-


[NH4+][OH- ]

[NH3]Kb = 1.8 x 10-5 =

(0.15) (x)(0.45)

1.8 x 10-5 =

So, x = [OH-] = _____________

and pH = _________________

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a) 0.0750 M pyridine, C5H5Nb) 0.0750 M C5H5N and 0.0850 M pyridinium chloride, C5H5NHClNote that C5H5NHCl,( salt) dissociates into C5H5NH+ and Cl-.

The Kb for pyridine is 1.7 x 10-9.


Answers: a) b)

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Buffers

Buffers, or buffered solutions, are special mixtures that are resistant to large pH changes, even when small amounts of strong acid or strong base are added.

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Buffers

· Buffers are able to resist large pH changes because they contain both an acidic component and a basic component.

· Buffers are prepared by mixing either:1) a weak acid and a salt containing its conjugate base OR2) a weak base and a salt containing its conjugate acid

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Buffers

Consider a buffer solution composed of HF and NaF.

· The acidic component is HF. This component aids in the neutralization of any strong base that is added to the buffer.

· The basic component is the fluoride ion, F-. This component aids in the neutralization of any strong acid that is added to the buffer.

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Buffers

If a small amount of KOH is added to this buffer, for example, the HF reacts with the OH− to make __________.

Consider a buffer solution composed of HF and NaF.

HF F-HF F-

F- HF

F- + H2O <-- HF + OH- H+ + F- --> HF

OH-H+

Buffer afteraddition of OH-

Buffer with equal conc. of weak acid and its conj. base

F-

HF

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Buffers

Similarly, if a small amount of strong acid (H+) is added, the F− reacts with it to form ____________.

HF F-HF F-F- HF

F- + H2O <-- HF + OH- H+ + F- --> HF

OH- H+

Buffer afteraddition of OH-

Buffer with equal conc. of weak acid and its conj. base

Buffer after addition of H+

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10 If a buffer is made of HA (weak acid) and NaA, which species will counteract the addition of a strong base?

A HAB Na+

C A-

D Both HA and Na+

E Both HA and A-

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11 If a buffer is made of HNO2 and KNO2 , which species will counteract the addition of a strong acid?

A HNO2

B K+

C NO2-

D Both HNO2 and K+

E Both K+ and NO2-

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Buffer capacity and pH

Two important characteristics of buffers:1- Its resistance to change in pH2- Its buffer capacity

Buffer capacity: The amount of acid or base the buffer can neutralize before the pH begins to change.

It depends on the amount of acid and base from which the buffer is made.

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[ H3O+] = Ka

The above two combination of solution will have the same [H+].

The pH will depend on ka and the relative concentration of acid and base only.

The buffering capacity will be higher for the first buffer. It contains greater number of moles of CH3COOH and NaCH3COO-.It can neutralize more of the acid/base added.

The greater the amounts, greater resistance to pH change.

Buffer capacity and pH

Buffer A Buffer B

1.0 M CH3COOH and1.0 M CH3COONa

Total volume = 1.0 L

0.1 M CH3COOH and0.1 M CH3COONa

Total volume = 1.0 L

Ka = [H3O+][C2H3O2- ]

[HC2H3O2]

[HC2H3O2][C2H3O2

- ]= 1Since

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12 Of the following solutions, which has the greatest buffering capacity?

A 0.1M NaF + 0.1MHF

B 0.5M NaF + 0.55M HF

C 0.8M NaF + 0.8M HF

D 0.2M NaF + 0.2M HF

E 1M NaF + 1M HF

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Buffer Calculations

Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

HA + H2O ↔ H3O+ + A-

Ka = [H3O+][A-]

[HA]

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Buffer CalculationsRearranging slightly, this becomes

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

Taking the negative log of both sides, we get

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

pKa = pH - log [base][acid]

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Buffer Calculations

Since pKa = pH - log [base][acid]

Rearranging this:pH = pKa + log [base]

[acid]

This is the Henderson-Hasselbalch equation

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Buffer calculations typically require you to calculate one or more of the following:

i) the pH of the buffer alone (Sample Prob #6)

ii) the pH of the buffer after a small amount of strong base has been added (and neutralized)(Sample Prob #7)

iii) the pH of the buffer after a small amount of strong acid has been added (and neutralized) (Sample Prob #8)

Because of the buffer system, the pH will not change drastically; it is usually less than a factor of 1.0 on the pH scale.

Buffer Calculations

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SAMPLE PROBLEM #6

What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 ´ 10−4.

Buffer Calculations

Method 1 - Traditional approachWrite the Ka expression. Solve for H+ and pH using ICE chart

HC3H5O3 (aq) + H2O(l) ↔ H3O+(aq) + C3H5O3-(aq)

0.12 0 0.10 -x +x +x 0.12-x x 0.10+x

1.4 ´ 10−4 = 0.10 x 0.12x = 0.000168PH =3.77

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SAMPLE PROBLEM #6What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 ´ 10−4.

Buffer Calculations

Method 2 -Using the Henderson-Hasselbalch equation

Henderson–Hasselbalch Equation

pH = pKa + log [base][acid]

pH = -log (1.4 x 10-4) + log (0.10)(0.12)

pH = 3.85 + (-0.08)

pH = 3.77

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13 The pH of a buffer solution that contains 0.818 M acetic acid (Ka = 1.76x10-5) and 0.172 M sodium acetate is __________.

A 4.077

B 5.434

C 8.571

D 8.370

E 9.922

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14 The pH of a buffer solution containing 0.145 M HF and 0.183 M KF is _____. (Ka of HF is 3.5x10-4)

A 3.56 B 2.19 C 2.66 D 3.35

E 4.32

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Recall that buffers resist drastic changes in pH, even when small amounts of either a strong acid or a strong base are added to it.

When this happens, we assume that all of the strong acid or strong base is consumed in the reaction. In other words, all of the strong acid or base gets completely neutralized by ONE of the components of the buffer system.

Addition of Strong Acid or Strong Base to a Buffer

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X- + H3O+ ⇒ HX + H2O

HX + OH- ⇒ X- + H2O

RecalculateHX and X-

Use Ka, [HX] and X-

to calculate [H+] pH

Add strong acid

Add strong base

stoichiometric calculation equilibrium calculation

Remember that all of the strong acid or base is consumed in the reaction.


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15 If a buffer is made of NH3 and NH4Cl, which component of the buffer will neutralize a small amount of HCl that is added?

A NH3

B NH4+

C Cl-

D Both A and B

E Both B and C

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16 If a buffer is made of HNO2 and KNO2 , which component of the buffer will neutralize a small amount of Ba(OH)2 that is added?

A HNO2

B K+

C NO2-

D Ba2+

E Both K+ and NO2-

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17 If a buffer is made of NH3 and NH4Cl, which component will neutralize any KOH that is added?

A NH3

B NH4+

C Cl-

D Both HNO2 and K+

E Both K+ and NO2-

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· Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.· Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

0.3M HC2H3O20.3M NaC2H3O2

add

0.02

M O

H-

add 0.02M H

+

pH =4.80

pH= 4.68

BufferpH= 4.74

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Calculating pH Changes in Buffers

SAMPLE PROBLEM #7 A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added. Assume volume change is negligible.

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Problem-solving strategy

Phase 1 - stoichiometric neutralization

· Identify the added substance. · Identify the component of the buffer that will neutralize the added substance.· Write down the equation for nutralization. · Solve for the new concentration of the buffer components.

Phase 2 - equilibrium · Write the relevant dissociation equation.· Create the ICE chart with the new [M] values(phase1) of acid and base components of the buffer.· Use Henderson-Hasselbalch equation to solve for pH.


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Phase 1 - Neutralization · strong acid or a strong base added ? · Which component of the buffer will neutralize the added substance?

SAMPLE PROBLEM #7 - AnswerA buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.


HA(aq)acid

OH-(aq)added

substanceH2O(aq) A−(aq)

base

Before 0.3 0.02 0.3

Neutralization -0.02 -0.02 0.3+0.02

After 0.28 0.0 0.32

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Phase 2 - equilibrium · Write the relevant dissociation equation.· Make a ICE chart using the amounts from Phase 1.

SAMPLE PROBLEM #7 - Answer (con't)A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.


HA(aq) H2O(l) H3O+(aq) A−(aq) Initial 0.28 M 0 0.32 M

Change -x +x +x

At Equilibrium 0.28-x x 0.32+x

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[H3O+][A−][HA]Ka =

Use the quantities from the ICE chart to calculate pH. You will need to look up the Ka value.

(x) 0.32)(0.28)1.8 x 10-5 =

So, x = [H3O+] = _______________

and pH = ___________

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Calculating pH Changes in BuffersAlternatively, you can use the Henderson-Hasselbalch equation to calculate the new pH:

pH = - log (1.8 x 10-5) + log (0.320)(0.280)

pH = 4.74 + 0.06

pH = 4.80

pH = pKa + log [base][acid]

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Calculating pH Changes in BuffersSAMPLE PROBLEM #8 A buffer is made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO to enough water to make 1.00 L of solution. Assume volume changes are negligible. Ka for cyanic acid = 3.5 x 10-4

a) Calculate the pH of the buffer.b) Calculate the pH of the buffer after the addition of 0.015 mol KOH.c) Calculate the pH of the buffer after the addition of 0.018 mol HNO3.

a) 3.351b) 3.46c) 3.22

Answers

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pH Range for Buffers

· The pH range is the range of pH values over which a buffer system works effectively.

· It is best to choose an acid with a pKa close to the desired pH.

· After studying titration graphs, you will see why buffers work best when pKa is close to the desired pH.

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SAMPLE PROBLEM #9 A buffer is made by adding 15.0 g ammonia, NH3, and 55.0 g ammonium chloride, NH4Cl. to enough water to make 1.00 L of solution. Kb for NH3 = 1.8 x 10-5. Assume volume changes are negligible.

a) Calculate the pH of the buffer.b) Calculate the pH of the buffer after the addition of 0.013 mol HClO4.c) Calculate the pH of the buffer after the addition of 0.015 mol KOH.

a) 9.193b) 9.181c) 9.206

Answers

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Calculating pH Changes in BuffersSAMPLE PROBLEM #10 A buffer is made by adding 25.0 g ammonia, NH3, and 45.0 g ammonium chloride, NH4Cl. to enough water to make 1.75 L of solution. Kb for NH3 = 1.8 x 10-5. Assume volume changes are negligible.

a) Calculate the pH of the buffer.b) Calculate the pH of the buffer after the addition of 0.011 mol NaOH.c) Calculate the pH of the buffer after the addition of 0.016 mol HNO3.

a) 9.50b) 9.51c) 9.49

Answers

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Titration

In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).

The concentration of the unknown will then be determined.

This is a quantitative analysis method.

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Titration

Rinse the buret with distilled waterRinse with the titrantFill the buret with the titrantDrain a small portion of the titrant so that air bubbles near the tip of the burete is expelled and filled to the tip.

Important things to remember in getting ready for titration:

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Rinse the pipette with distilled water

Rinse with the solution to be pipeted

Pipette the solution , adjust the level and dispense the fixed volume to a beaker or flask.

TitrationImportant things to remember in getting ready for titration:

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TitrationA pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

At the equivalence point,

# of moles of acid = # of moles of base

MaVa = MbVb

By using an indicator you could visually see the sudden change in color of the indicator at this point. This point is also known as the "end point" where you will stop adding the titrant.

The end point is the appearance of the first permanent color change of the indicator.

Note that this point will have a slight excess of the titrant in the solution

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At the equivalence point, ( stoichiometric)


MaVa = MbVb

By using an indicator you could visually see the sudden change in color of the indicator at this point. This point is also known as the "end point" where you will stop adding the titrant.

The end point is the appearance of the first permanent color change of the indicator.

Note that this point will have a slight excess of the titrant in the solution

This minute amount of the titrant is making the indicator color visible.

TitrationEquivalence point and End point

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At the equivalence point,


MaVa = MbVb

This equation is only applicable for reactions in which the ratio of

moles acid to moles base is 1:1.

Titration - Equivalence Point

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HCl + NaOH --> NaCl + H2OHBr + KOH --> KBr + H2O

HNO3 + LiOH --> LiNO3 + H2OCH3COOH + NaOH --> CH3COONa + H2O

So for reactions such as these, use the equation above to calculate the molarity or volume of acid or base at the equivalence point.

MaVa = MbVb

Here are some examples of reactions in which the ratio of moles acid to moles base is 1:1.


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2 HCl + Ca(OH)2 --> CaCl2 + 2 H2O2 HBr + Sr(OH)2 --> SrBr2 + 2 H2O

2 HNO3 + Ba(OH)2 --> Ba(NO3)2 + 2 H2O

Consider some examples of reactions in which the ratio of moles acid to moles base is 2:1.

moles acidmoles base

21= MaVa

MbVb=

Cross-multiplying, we obtain MaVa = 2 MbVb

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H2SO4 + 2 KOH --> K2SO4 + 2 H2OH2SO3 + 2 LiOH --> Li2SO3 + 2 H2O

H2C2O4 + 2 NaOH --> Na2C2O4 + 2 H2O

Consider some examples of reactions in which the ratio of moles acid to moles base is 1:2.

moles acidmoles base

12=

Cross-multiplying, we obtain 2 MaVa = MbVb

MaVaMbVb

=


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TitrationSummary of Equations for

Equivalence Point

Equation Acid Base

MaVa = MbVb HCl, HNO3, HBr, CH3COOH, HClO4

NaOH, KOH, LiOH

2 MaVa = MbVb H2SO4, H2SO3, H2C2O4

NaOH, KOH, LiOH

MaVa = 2 MbVb HCl, HNO3, HBr, CH3COOH, HClO4

Ca(OH)2, Sr(OH)2,Ba(OH)2

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18 What is the concentration of hydrochloric acid if 20.0 mL of it is neutralized by 40mL of 0.10M sodium hydroxide?

A 0.025 M

B 0.050 M

C 0.10 M

D 0.20 MMaVa = MbVb

2 MaVa = MbVb

MaVa = 2 MbVb

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19 What is the concentration of KOH if 60 mL of it is neutralized by 20 mL 0.10M HCl?

A 0.005 MB 0.025 MC 0.033 MD 0.10 ME 0.30 M

MaVa = MbVb

2 MaVa = MbVb

MaVa = 2 MbVb

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20 What is the concentration of sulfuric acid if 50mL of it is neutralized by 10mL of 0.1M sodium hydroxide?

A 0.005M

B 0.01M

C 0.25M

D 0.5MMaVa = MbVb

2 MaVa = MbVb

MaVa = 2 MbVb

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21 How much 0.5 M HNO3 is necessary to titrate 25 mL of 0.05M Ca(OH)2 solution to the endpoint?

MaVa = MbVb

2 MaVa = MbVb

MaVa = 2 MbVb

A 2.5 mL

B 5.0 mL

C 10 mL

D 20 mL

E 25 mL

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22 How much 3.0 M NaOH is needed to exactly neutralize 20.0 mL of 2.5 mL H2SO3?

A 8.3 mL

B 17 mL

C 24 mL

D 33 mLE 48 mL

MaVa = MbVb

2 MaVa = MbVb

MaVa = 2 MbVb

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23 How much 1.5 M NaOH is necessary to exactly neutralize 20.0 mL of 2.5 M H3PO4?

MaVa = MbVb

2 MaVa = MbVb

MaVa = 2 MbVb

A 11 mLB 12 mLC 33 mLD 36 mLE 100 mL

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24 What is the molarity of a NaOH solution if 15 mL is exactly neutralized by 7.5 mL of a 0.02 M HC2H3O2 solution?

MaVa = MbVb

2 MaVa = MbVb

MaVa = 2 MbVb

A 0.005 M

B 0.010 M

C 0.020 M

D 0.040 M

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Typically, there are 4 "zones" in which you may be asked to calculate pH:

· Before any titrant is added from the buret

· After a small amount of titrant has been added

· At the equivalence point

· After the equivalence point

First, we will examine these zones on a titration graph. Then, we will review the pH calculation for each region.

The strategy for calculating pH is different for each zone.

Titration of a Strong Acid with a Strong Base

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From the start of the titration to near the equivalence point, the pH goes up slowly.

The low initial pH indicates that the substance being titrated is a strong acid.

(1) Before any titrant is

added from the buret

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Just before (and after) the equivalence point, the pH increases rapidly.

(2) After a small amount of titrant has been added*

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At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

(3) At the equivalence

point

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As more base is added, the increase in pH again levels off.

(4) After the equivalence

point

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In summary, the four regions of any titration graph are:

· Before any titrant is added from the buret

· After a small amount of titrant has been added

· At the equivalence point

· After the equivalence point

The pH calculation differs for each "zone," so we will consider each one separately.


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Solving Titration Problems: Strong Acid - Strong Base

1) Before any titrant is added from the buret

The pH depends up on the concentration of the acid or base.

Use the molarity of the acid or base to determine the pH.pH = - log [H3O+] OR pH = 14 - (-log [OH-]

Remember to check the acid or base is polyprotic or polyhydroxy.For example: · 20 ml 0.5M HCl is titrated with 0.25M NaOHThe original acid is 0.5M; The concentration of H+ = 0.5MMaVa = MbVb

· 20 ml 0.5M H2SO4 is titrated with 0.25M NaOHThe original acid is 0.5M; The concentration of H+ = 0.5x2 M2MaVa = MbVb

20 ml 0.5M HCl is titrated with 0.25M Ca(OH)2The original acid is 0.5M; The concentration of H+ = 0.5 MMaVa = 2MbVb2

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2) When some titrant is added from the buret, before the equivalence point is reachedExample: 20 ml 0.5M HCl is titrated with 0.25M NaOH. What is the pH after 10 ml of base is added?

· It is recommended that you calculate the volume of the titrant needed to neutralize the acid at the beginning to see which segment of the titration we are in.

write down the neutralization reaction

HCl + NaOH --> NaCl + H2O0.01 mol 0.0025mol

-0.0025mol -0.0025mol

0.0075mol 0 mol left

[H+] after the 10 ml base had been neutralized by the acid is = 0.0075 mols /0.030L = 0.25MpH = 0.6


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3) At the equivalence point: Example: 20 ml 0.5M HCl is titrated with 0.25M NaOH. What is the pH at the equivalence point ?

· You will know you have reached the equivalence point when MOLES of acid = MOLES of base

· Since there is no excess [H+] or [OH-], we must look to the salt that is formed to see if it will affect pH. · Since it is from a strong base and strong acid, will not undergo hydrolysis to change the pH.

· In cases such as these, when a strong acid and strong base are titrated, the pure water will have a pH = 7.0

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4) Beyond the equivalence point, when excess titrant has been added from the buret

Example: 20 ml 0.5M HCl is titrated with 0.25M NaOH. What is the pH after 45 ml of base had been added?

0.5 x 20ml = 0.25 x V2mlV2 = 40 ml NaOH need to neutralize the 20 ml acid.

write down the neutralization reaction

HCl + NaOH --> NaCl + H2O0.01 mol 0.0125mol

-0.01mol -0.01mol

0 mol 0.0025mol excess

[OH-] after the 45 ml base added = = 0.0025 mols /0.065L = 0.0385MpOH = 1.415pH = 12.585

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The pH at eqequivalence point will be =7Acid in the flask and titrant is the base.

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Acid in the flask and base is the titrant

Base in the flask and acid is the titrant

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25 The moles of acid equals the moles of base at the equivalence point in a titration of _____ with _____.A strong acid, weak base

B strong base, weak acid

C strong acid, strong base

D weak acid, weak base

E All of the above

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26 What is the pH of a titration between a weak acid and a strong base at the equivalence point?

A Less than 7

B Equal to 7

C Greater than 7

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27 What is the pH of a titration between a weak base and a strong acid at the equivalence point?

A Less than 7

B Equal to 7

C Greater than 7

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Solving Titration Problems: Weak - Strong

1) Before any titrant is added from the buret

Consider the dissociation equation for the substance in the flask:

For a weak acid in the flask For a weak base in the flaskKa = x2 / [acid] Kb = x2 / [base]pH = - log x pH = 14 - (-log x)

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2) When some titrant is added from the buret, before the equivalence point is reachedexample# 11: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH; 10 ml NaOH is added: ka = 1.8 x 10-5

CH3COOH + NaOH --> Na+CH3COO- + H2O 0.01mol 0.0025mol-0.0025mol -0.0025mol + 0.0025mol0.0075mol 0.0 0.0025mol

0.25M (acid) 0.083M ( c.base)

Remember you have a buffer situation now because of the salt produced from the neutralizationof the weak acid.

Use the Henderson-Hasselbalch equation to solve for the pH of the solution at this point.

pH = pKa + log ( [base] / [acid] )

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Sample # 12calculate the pH of the solution after adding 10 ml of 0.05M KOH to 40 ml of 0.025M Benzoic acid. Ka for benzoic acid = 6.3 x 10-5


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sample #13Calculate the pH of the solution after adding 10 ml of 0.1M HCl to 20 ml of 0.1 M NH3Kb of NH3 = 1.8 x 10-5

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3) At the equivalence pointexample # 14: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH; 40 ml NaOH is added: ka = 1.8 x 10-5

CH3COOH + NaOH --> Na+ CH3COO- + H2O 0.01mol 0.01mol-0.01mol -0.01mol + 0.01mol0.0mol 0.0 mol 0.01mol

0.166M (base)The salt Na+ CH3COO- has a strong conjugate base CH3COO-

The anion, CH3COO- will undergo hyrolysis as follows:

CH3COO- + H2O --> CH3COOH + OH-

0.166M 0 0-X +x +x0.166-X x x

Kb = X2 / 0.166 ; Solve for x Remember Kb = Kw / KaCalculate pHWill be slightly basic !

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sample # 15Calculate the pH at equivalence point when 40 ml of 0.025M benzoic acid is titrated with 0.05M KOH. Ka of benzoic acid is 6.3 10-5

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Sample # 16calculate the pH at the equivalence point when 40 ml of 0.1M NH3 is titrated with 0.2 M HCl.Kb of NH3 is 1.8 10-5

Slide 108 / 113Titration of a Weak Acid with a

Strong Base

The pH at the equivalence point is above 7

weak acid vs strong base

strong acid vs strong base

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Titration of a Weak Acid with a Strong Base

With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

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Titration of a Weak Base with a Strong Acid

· The pH at the equivalence point in these titrations is < 7.· Methyl red is the indicator of choice.

strong base with strong acid , pH =7weak base NH3 with strong acid , pH = 5.5

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4) Beyond the equivalence point, when excess titrant has been added from the buretexample: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH; 50 ml NaOH is added:

This is exactly the same as for strong acid-strong base titrations.

The strong base added after the end point will increase the OH- in the solution making it strongly basic.

Calculate the [OH-] from the excess base in the new volume and determine the pH.

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Titrations of Polyprotic Acids

When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.

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