Buffers/Titration Review hydrolysis of salts Aqueous Equilibria - I - - PDF document

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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

SLIDE 2

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5 A basic solution would result from the hydrolysis of one

• f 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 Common-Ion Effect

“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

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

The Common-Ion Effect

HF(aq) H2O(l) H3O+(aq) F−(aq)

Initial 0.20 M 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

The Common-Ion Effect

SLIDE 3

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The Common-Ion Effect

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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The Common-Ion Effect

HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq) 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 0.10 M

Change at Equilibrium

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HF(aq) H2O(l) H3O+(aq) F−(aq)

Initial 0.20 M 0.10 M Change

• x

+x +x At Equilibrium 0.20-x x 0.10 + x

The Common-Ion Effect

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The Common-Ion Effect

Therefore, x = [H3O+] = 1.4 x 10−3 and pH = −log (1.4 x 10−3) pH = 2.87

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The Common-Ion Effect

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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HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq) · Consider the two solutions we just examined in Sample Problem #1. · Compare the following: · How do these results support Le Chatelier's Principle?

The Common-Ion Effect

Solution Final [H3O+] pH HF 0.12 HF and NaF 1.4 x 10-3

SLIDE 4

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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.

The Common-Ion Effect

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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.

The Common-Ion Effect

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+].

The Common-Ion Effect

[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?

The Common-Ion Effect

Solution [OH-] pH HC2H3O2 HC2H3O2 and NaC2H3O2

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SAMPLE PROBLEM #3 Calculate the pH of the following solutions: a) 0.85 M nitrous acid, HNO2 b) 0.85 M HNO2 and 0.10 M potassium nitrite, KNO2 The Ka for nitrous acid is 4.5 x 10-4.

The Common-Ion Effect

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

SLIDE 5

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

A

KOH

B

AlCl3

C

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

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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The Common-Ion Effect

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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SAMPLE PROBLEM #4 Calculate the pH of the following solutions:

a) a 0.45 M solution of NH3 b) a 0.45 M solution of NH3 that also contains 0.15 M NH4Cl

The Kb for NH3 is 1.8 x 10-5.

The Common-Ion Effect

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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-

The Common-Ion Effect

[NH4

+][OH- ]

[NH3] Kb = 1.8 x 10-5 = x = [OH-] = _____________

pH = _________________

x2 (0.45) 1.8 x 10-5 =

SLIDE 6

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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-

The Common-Ion Effect

[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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SAMPLE PROBLEM #5 Calculate the pH of the following solutions:

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

The Kb for pyridine is 1.7 x 10-9.

The Common-Ion Effect

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 OR 2) 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 after addition of OH- Buffer with equal

• conc. of weak acid

and its conj. base

F

• HF

SLIDE 7

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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 after addition 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

HA

B

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 pH 2- 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 and 1.0 M CH3COONa Total volume = 1.0 L 0.1 M CH3COOH and 0.1 M CH3COONa Total volume = 1.0 L

Ka = [H3O+][C2H3O2

• ]

[HC2H3O2] [HC2H3O2] [C2H3O2

• ]

= 1 Since

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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

SLIDE 8

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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 Calculations

Rearranging 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 approach Write 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 0.12 x = 0.000168 PH =3.77

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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 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

SLIDE 9

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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 Recalculate HX 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.

Addition of Strong Acid or Strong Base to a Buffer

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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

SLIDE 10

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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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Addition of Strong Acid or Strong Base to a Buffer

· 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 HC2H3O2 0.3M NaC2H3O2 a d d . 2 M O H

• add 0.02M H+

pH =4.80 pH= 4.68

Buffer

pH= 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.

Calculating pH Changes in Buffers

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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 - Answer 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

• f the buffer is 4.74. Calculate the pH of this solution after

0.020 mol of NaOH is added.

Calculating pH Changes in Buffers

HA(aq) acid OH-(aq) added substance H2O(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

• f NaOH is added.

Calculating pH Changes in Buffers

HA(aq) H2O(l) H3O+(aq) A−(aq)

Initial 0.28 M 0.32 M Change

• x

+x +x At Equilibrium 0.28-x x 0.32+x

SLIDE 11

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Calculating pH Changes in Buffers

[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 = ___________ Slide 62 / 113

Calculating pH Changes in Buffers

Alternatively, 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 Buffers

SAMPLE 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

• f 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) 3.351 b) 3.46 c) 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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Calculating pH Changes in Buffers

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) 9.193 b) 9.181 c) 9.206

Answers

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Calculating pH Changes in Buffers

SAMPLE 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) 9.50 b) 9.51 c) 9.49 Answers

SLIDE 12

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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 water Rinse with the titrant Fill the buret with the titrant Drain 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.

Titration

Important things to remember in getting ready for titration:

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Titration

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

Titration

Equivalence point and End point

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SLIDE 13

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At the equivalence point, # of moles of acid = # of moles of base 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 + H2O HBr + KOH --> KBr + H2O HNO3 + LiOH --> LiNO3 + H2O CH3COOH + 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.

Titration - Equivalence Point

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Titration - Equivalence Point

2 HCl + Ca(OH)2 --> CaCl2 + 2 H2O 2 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 acid moles base 2 1 = MaVa MbVb = Cross-multiplying, we obtain MaVa = 2 MbVb

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H2SO4 + 2 KOH --> K2SO4 + 2 H2O H2SO3 + 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 acid moles base 1 2 = Cross-multiplying, we obtain 2 MaVa = MbVb MaVa MbVb =

Titration - Equivalence Point

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Titration

Summary 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 M MaVa = MbVb 2 MaVa = MbVb MaVa = 2 MbVb

SLIDE 14

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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 M

B

0.025 M

C

0.033 M

D

0.10 M

E

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

• f it is neutralized by 10mL of 0.1M sodium

hydroxide?

A

0.005M

B

0.01M

C

0.25M

D

0.5M MaVa = 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 mL

E

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 mL

B

12 mL

C

33 mL

D

36 mL

E

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

SLIDE 15

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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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Titration of a Strong Acid with a Strong Base

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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Titration of a Strong Acid with a Strong Base

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

(2) After a small amount

• f titrant has

been added*

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Titration of a Strong Acid with a Strong Base

At the equivalence point, moles acid = moles base, and the solution contains

• nly 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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Titration of a Strong Acid with a Strong Base

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

(4) After the equivalence point

SLIDE 16

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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.

Titration of a Strong Acid with a Strong Base

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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 NaOH The original acid is 0.5M; The concentration of H+ = 0.5M MaVa = MbVb · 20 ml 0.5M H2SO4 is titrated with 0.25M NaOH The original acid is 0.5M; The concentration of H+ = 0.5x2 M 2MaVa = MbVb 20 ml 0.5M HCl is titrated with 0.25M Ca(OH)2 The original acid is 0.5M; The concentration of H+ = 0.5 M MaVa = 2MbVb2

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2) When some titrant is added from the buret, before the equivalence point is reached Example: 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 + H2O 0.01 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 = 0.25M pH = 0.6

Solving Titration Problems: Strong Acid - Strong Base

Slide 94 / 113 Solving Titration Problems: Strong Acid - Strong Base

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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Solving Titration Problems: Strong Acid - Strong Base

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 V2ml V2 = 40 ml NaOH need to neutralize the 20 ml acid. write down the neutralization reaction HCl + NaOH --> NaCl + H2O 0.01 mol 0.0125mol

• 0.01mol -0.01mol

0 mol 0.0025mol excess [OH-] after the 45 ml base added = = 0.0025 mols /0.065L = 0.0385M pOH = 1.415 pH = 12.585

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Solving Titration Problems: Strong Acid - Strong Base

The pH at eqequivalence point will be =7

Acid in the flask and titrant is the base.

SLIDE 17

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Solving Titration Problems: Strong Acid - Strong Base

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

Slide 101 / 113 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 flask Ka = x2 / [acid] Kb = x2 / [base] pH = - log x pH = 14 - (-log x)

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Solving Titration Problems: Weak - Strong

2) When some titrant is added from the buret, before the equivalence point is reached example# 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 + 0.0025mol

0.0075mol 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] )

SLIDE 18

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Sample # 12 calculate 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

Solving Titration Problems: Weak - Strong Slide 104 / 113 Solving Titration Problems: Weak - Strong

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

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Solving Titration Problems: Weak - Strong

3) At the equivalence point example # 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 + 0.01mol

0.0mol 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 +x

0.166-X x x Kb = X2 / 0.166 ; Solve for x Remember Kb = Kw / Ka Calculate pH Will be slightly basic !

Slide 106 / 113 Solving Titration Problems: Weak - Strong

sample # 15 Calculate 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 # 16 calculate 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

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Titration 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

SLIDE 19

Slide 109 / 113 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 =7 weak base NH3 with strong acid , pH = 5.5

Slide 111 / 113 Solving Titration Problems: Weak - Strong

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