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York University CHEM 1001 3.0 Acids and Bases - 48

Buffers and Acid-Base Titrations

Reading: from chapter 18 of Petrucci, Harwood and Herring(8th edition):

Required: Sections 18-2 through 18-6.

Recommended: Section 18-1.

Examples: 18-3 through 18-9.

Problem Set:

Chapter 18 questions: 5, 6, 7, 8, 9, 26, 27a-e, 32.

Additional problems from Chapter 18: 54, 55.


Adding Base to a Weak Acid SolutionHF is a weak acid; it partially ionizes in water:

HF + H2O W H3O+ + F- pKa = 3.18

What effect does adding NaOH have on this equilibrium?

NaOH 6 Na+ + OH- complete ionization

H3O+ + OH- 6 H2O neutralization

Consequences:

C Adding NaOH consumes H3O+.

C The HF equilibrium shifts to the right to partiallyoffset the change (replaces some H3O

+).

C The increase in pH is less than if the same amount ofNaOH was added to pure water.


Adding Acid to a Salt of a Weak AcidNaF is a salt of a weak acid (HF). It hydrolyzes in water:

F- + H2O W OH- + HF pKb = 10.82

What effect does adding HCl have on this equilibrium?

HCl + H2O 6 H3O+ + Cl- complete ionization

H3O+ + OH- 6 H2O neutralization

Consequences:

C Adding HCl consumes OH-.

C The hydrolysis equilibrium shifts to the right to partiallyoffset the change (replaces some OH-).

C The decrease in pH is less than if the same amount ofHCl was added to pure water.


Solution of a Weak Acid and its Salt

Example: Solution of HF and NaF in water.

HF + H2O W H3O+ + F- pKa = 3.18

C Adding acid shifts the equilibrium to the left. Thispartially offsets the reduction in pH.

C Adding base shifts the equilibrium to the right. Thispartially offsets the reduction in pH.

C Significant amounts of acid or base can be neutralizedsince the solution contains substantial [HF] and [F-].

C The pH is kept nearly constant. This solution is a buffer.


Applications of Buffer Solutions

Many chemical processes require control of pH.

C The rates and mechanisms of many chemical reactionsare sensitive to pH.

C pH affects the structure of and charges on proteins.

C The activity of enzymes is very sensitive to pH.

C Human blood is buffered to a pH of 7.4

C Many industrial chemical processes require acontrolled pH.


Acid-Base Buffers

An acid-base buffer is a solution that is resistant tochanges in pH. In other words, when an acid or base isadded to a buffer, the pH changes only slightly.

For a solution to be a buffer, it must contain:

C comparable concentrations of a weak acid its conjugatebase:

0.1×[acid] < [conjugate base] < 10×[acid]

C a concentration of the conjugate base at least 100 timesthe value of Ka for the acid:

[conjugate base] > 100 Ka


pH of a Buffer SolutionReaction HA + H2O W H3O

+ + A-

Initial [HA]0 - 0 [A-]0 M

Change -x - x x M

Equil. [HA]0-x - x [A-]0+x M

Simplifying assumptions: x n [HA]0 and x n [A-]0

Y Ka = [H3O+][A-]/[HA] . x [A-]0 / [HA]0

Y [H3O+] = x = Ka [HA]0 / [A

-]0

Check that the assumptions are valid:

C Since [A-]0 > 100Ka, x < 0.01[HA]0.

C Since [A-]0 > 100Ka and [HA]0 < 10[A-]0, x < 0.1[A-]0.


pH of a Buffer Solution - continued

In a buffer, we have:

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

Take the negative logarithm of both sides:

-log[H3O+] = -logKa - log([HA]/[A-])

pHbuffer = pKa + log([A-]/[HA])

This is called the Henderson-Hasselbalch equation.

C Relates the pH of a buffer to its composition.

C Uses initial concentrations of acid and conjugate base.

C Don't memorize - it is easy to mess up but easy to derive.


Buffer RangeTo have a buffer, we must have:

0.1 < [A-] / [HA] < 10

Taking logarithms:

-1.0 < log([A-]/[HA]) < 1.0

Substitute into the Henderson-Hasselbalch equation:

pHbuffer = pKa + log([A-]/[HA])

pHbuffer = pKa ± 1.0

Conclusion: A particular weak acid can used to preparebuffers with pH values within 1 unit of the acid's pKa.


Preparing Buffer Solutions

Chemists and biologists often need to prepare solutionswith a specific, constant pH.

To make a buffer with a specific pH:

C Find an acid with a pKa within 1 unit of the required pH.

C Use the acid ionization constant to determine therequired ratio of acid to conjugate base.

C Make the buffer solution.

C Use a pH meter to check the pH. Add acid or base asneeded to get the correct pH.


Acid pKa

Citric Acid 3.13

Benzoic acid 4.20

Acetic acid 4.77

Carbonic acid 6.36

Ammonium ion 9.25

Phenol 9.89

Preparing a Buffer - Example

To prepare a buffer with pH = 3.50:

» Closest to desired pH.

Find ratio of acid to base:

[HA]/[A-] = [H3O+]/Ka = 10pKA-pH

[HA]/[A-] = 103.13-3.50 = 0.43

if [A-] = 0.10 M (>100Ka)then [HA] = 0.043 M.

For one liter, use 0.143 mol H3C6H5O7 plus 0.100 mol NaOH.


Adding Acids or Bases to a Buffer

Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

C Adding an acid decreases [A-] and increases [HA]. So itlowers the pH.

C Adding a base decreases [A-] and increases [HA]. So itraises the pH.

C Qualitatively, this is just like any other solution. But ina buffer, the pH changes are much smaller.

Q: How much does the pH change when acid or base is added?

A: Calculate the changes in [HA] and [A-].


pH Change in a Buffer

Given a citric acid-citrate buffer with [A-]0 = 0.100 M,[HA]0 = 0.043 M, and pH = 3.50 ([H3O

+] = 3.16×10-4 M).

Find the pH after adding 3.0×10-3 mol HCl to 1.0 liter ofthis buffer.

Solution: Calculate the changes in [HA] and [A-].

[A-] = [A-]0 - 3.0×10-3 M = (0.100-0.003) M = 0.097 M

[HA] = [HA]0 + 3.0×10-3 M = (0.043+0.003) M = 0.046 M

pH = pKa + log([A-]/[HA]) = 3.13 + log(0.097/0.046)

pH = 3.45 Y [H3O+] = 3.55×10-4 M

The change in [H3O+] equals 1.3% of the added HCl.


Effect of Buffer Concentration

Given a citric acid-citrate buffer with [A-]0 = 0.100 M,[HA]0 = 0.043 M, and pH = 3.50.

What happens to the pH if we double both [A-]0 and [HA]0?

Answer: No change since pH = pKa + log([A-]/[HA]).

(Only approximately true since we really should be usingactivities instead of concentrations.)

However, the buffer capacity will increase, therefore:

C The effect of added acid or base will be smaller.

C More acid or base could be added before the bufferstops working.


Buffer Solutions - Summary

C A buffer solution contains a weak acid and the conjugatebase of that acid in comparable amounts.

C The acid neutralizes added base and the conjugate baseneutralizes added acid.

C A buffer dramatically reduces the pH changes producedby adding an acid or base.

C The pH of a buffer is close to the pKa of the weak acid.

C The pH of a buffer solution is readily calculated from theequilibrium condition

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


Determining pHTwo widely used methods:

C A pH meter uses an electrode that produces a voltagethat is a linear in pH.

C Acid-base indicators change color depending on pH.An indicator is a weak acid that has a different colorthan its conjugate base.

HIn + H2O W H3O+ + In-

acid color base color

Acid color dominates for pH < pKa - 1.Base color dominates for pH > pKa + 1.

Easy to use, but results are less precise.


Acid-Base Indicator Examples

pH =

Phenolphthalein

pKa = 9.1

pH =

Bromthymol blue

pKa = 7.1


Acid-Base Titrations

Acid-base titrations are used to analyze the quantity of acidor base in an unknown sample.

Acid + Base W H2O + salt

At the equivalence point:

moles of acid = moles of base Y CAVA = CBVB

VA = volume of acid of unknown concentration CA

CB = known concentration of the base

VB = volume of base added at equivalence.

Y CA = CBVB / VA


Acid-Base Titrations - continued

C The titrant is the strong acid or strong base solution that isadded in a measured volume.

C The equivalence point can be detected by using a pH meter.

C The end point is an experimental approximation of theequivalence point, identified with an acid-base indicator.

C The titration error is the difference between theequivalence point and the end point.

How close is the end point to the equivalence point?

What indicator should be used?


Titration of 25.00 ml of 0.100 M HCl with 0.100 M NaOH.

Titration Curve of a Strong Acid


pH during Titration of a Strong Acid

During the titration of a strong acid with a strong base:

C Initial pH is determined by the concentration of the acid.

C During the titration, the acid is gradually neutralized.Before the equivalence point:

nA = CAVA, nB = CBVB, [H3O+] = (nA-nB) / (VA+VB)

C At the equivalence point: pH = 7.

C Beyond the equivalence point:

[OH-] = (nA-nB) / (VA+VB)

C What about near the equivalence point?


pH near the Equivalence Point Titration of 25.00 ml of 0.100 M HCl with 0.100 M NaOH.

At the equivalence point: VB = 25.00 ml and pH = 7.

Find the pH at VB = 24.95 ml (1 drop short of equivalence).

Solution:

nA = (25.00 ml)(0.100 M) = 25.00 mmol

nB = (24.95 ml)(0.100 M) = 24.95 mmol

[H3O+] = (0.05 mmol) / (49.95 ml) = 0.0001 M

Y pH = 4

By same method, one drop past equivalence, pH = 10.

Conclusion: Near equivalence, pH changes very rapidly.


Titration of a Weak Acid

C The titrant must be a strong base.

C The initial pH is determined by the concentration andpKa of the acid.

C During the titration, the acid is gradually converted toits conjugate base. The solution is a buffer.

C At the equivalence point, the solution contains only asalt of the conjugate base, so pH >7.

C There is a less dramatic pH change near the equivalencepoint. Indicators must be chosen with care.


25.0 ml of 0.100 M HC2H3O2 titrated with 0.100 M NaOH.

Titration Curve of a Weak Acid


pH during Titration of a Weak Acid

Titrating 25.00 ml of 0.100 M acetic acid (HC2H3O2,pKa = 4.74) with 0.100 M NaOH.

C Initial pH:

HA + H2O W H3O+ + A-

[H3O+] = [A-] = x [HA] . CA = 0.100 M

x2 = Ka [HA] Y x = 1.35×10-3 M Y pH = 2.87

C Buffer region (10% to 90% neutralized):

pH = pKa + log([A-]/[HA]) = pKa + log(nB/(nA-nB))

where nA = CAVA, nB = CBVB

C At the half-equivalence point: pH = pKa = 4.74


Titration of a Weak Acid - continued C At the equivalence point:

The solution is that of a salt of the conjugate base.This hydrolyzes:

A- + H2O W HA + OH-

Kb = KW/Ka = [HA][OH-]/[A-] = x2/[A-]

[A-] . nA/(VA+VB) = CACB/(CA+CB) = 0.0500 M

x2 = Kb[A-] Y x = 5.24×10-6 M Y pOH = 5.28

Y pH = 8.72 Suitable indicators: thymol blue (pKa=8.8)phenolphthalein (pKa=9.1)

C Beyond the equivalence point, the pH is determined bythe amount of excess base.


Titrating 10.0 ml of 0.100 M H3PO4 with 0.100 M NaOH.

Titration Curve of a Weak Polyprotic Acid


pH during Titration of a Polyprotic Acid

C Initial pH: Same as for a monoprotic acid.

C Buffer regions: There are two or more, with pH.pKa1,pH.pKa2, etc.

C Last equivalence point: Same as for a monoprotic acid.

C Earlier equivalence points: Not so simple.

Example: For H3PO4 at its first equivalence point, thereare significant amounts of H3PO4, H2PO4

-, and HPO42-.

H3PO4 + H2O W H3O+ + H2PO4

- pKa1 = 2.15

H2PO4- + H2O W H3O

+ + HPO42- pKa2 = 7.20

Expect pKa1 < pH < pKa1 (small [H3O+] and [OH-])


Equivalence Point of a Polyprotic Acid

Find the pH at the first equivalence point of H3PO4.

Solution: Two equilibria - use the general method.

Significant species: H3PO4, H2PO4-, HPO4

2-, H3O+ (pH<6)

Material balance: [H3PO4] + [H2PO4-] + [HPO4

2-] / M

Equilibrium constants:

[H2PO4-] = Ka1[H3PO4]/[H3O

+]

[HPO42-] = Ka2[H2PO4

-]/[H3O+] = Ka1Ka2[H3PO4]/[H3O

+]2

Electroneutrality: [Na+] + [H3O+] = [H2PO4

-] + 2[HPO42-]

At equivalence: M = [Na+] . [H2PO4-] + 2[HPO4

2-]

Four equations, four unknowns.


Polyprotic Acid - continued

Substitute into material balance:

M = [H3PO4]( 1 + Ka1/[H3O+] + Ka1Ka2/[H3O

+]2 )

Substitute into electroneutrality:

M = [H3PO4]( Ka1/[H3O+] + 2Ka1Ka2/[H3O

+]2 )

Subtract electroneutrality from material balance:

0 = 1 - Ka1Ka2/[H3O+]2 Y [H3O

+] = (Ka1Ka2)½

Y pH = (pKa1+pKa2)/2 = (2.15+7.20)/2

pH = 4.68


Acid-Base Titrations - Summary

C Acid-base titrations can be used to measure theconcentrations of acid or base in solution.

C The pH changes rapidly at the equivalence point.

C The equivalence point can be determined by using apH meter or estimated by using an acid-base indicator.

C The appropriate indicator to use depends on the pKa (or pKb)of the acid (or base) being titrated.

C A titration curve can be used to determine the pKa of a weakacid (or pKb of a weak base).


CO2 in the Atmosphere and Oceans

The partial pressure of CO2 in the atmosphere is increasing.

C Pre-industrial: 2.8×10-4 atm

C Present: 3.7×10-4 atm

C Future: doubled? tripled? more?

CO2 can dissolve in water and act as an acid:

CO2(g) W CO2(aq) Keq = 0.046

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

- pKa1 = 6.05

HCO3- + H2O(l) W H3O

+ + CO32- pKa2 = 9.23

These equilibrium constants are for seawater at 15 °C.

What is the impact of increased CO2 on the oceans?


Why the Sea is Salt

Rainwater is slightly acidic, pH - 5.

Rainwater gradually dissolves rocks (mostly as ions).Dissolved material is washed into the oceans.

Water evaporates from the oceans, dissolved material isleft behind along with primordial Na+ and volcanic Cl-.

Composition of seawater:

C Major cations: Na+, K+, Mg2+, Ca2+

C Major anions: Cl-, SO42-, HCO3

-, Br-, CO32-

C The atmosphere directly influences HCO3- and CO3

2- inseawater. pH is a key factor.


pH of Seawater

Significant species: CO2(aq), HCO3-, CO3

2-, H3O+

Electroneutrality: sum of '+' charges = sum of '-' charges

[Na+] + [K+] + ... = [Cl-] + ... + [HCO3-] + 2[CO3

2-]

[Na+], [K+], [Cl-], etc. are constant, so [HCO3-]+2[CO3

2-]is a constant called the alkalinity:

[Alk] / [HCO3-] + 2[CO3

2-]

Equilibrium conditions: [CO2(aq)] = Keq PCO2

[HCO3-] = Ka1[CO2(aq)][H3O

+]-1 = Ka1KeqPCO2[H3O+]-1

[CO32-] = Ka2[HCO3

-][H3O+]-1 = Ka1Ka2KeqPCO2[H3O

+]-2

Four equations, four unknowns (PCO2 is assumed known).


pH of Seawater - continuedSubstitute into expression for alkalinity:

[Alk] = Ka1KeqPCO2[H3O+]-1 + 2Ka1Ka2KeqPCO2[H3O

+]-2

[H3O+] = Ka1KeqPCO2 (1 + 2Ka2/[H3O

+]) / [Alk]

For seawater: [Alk] . 2.0×10-3 M. Constants given earlier.

Simplifying assumption: [H3O+] o 2Ka2 = 1.2×10-9

Y [H3O+]approx = Ka1KeqPCO2 / [Alk] = (2.05×10-5 atm-1)PCO2

Results PCO2 (atm) [H3O+]approx [H3O

+]iterate pH

pre-industrial 2.8×10-4 5.7×10-9 M 6.7×10-9 M 8.17

doubled CO2 5.6×10-4 1.1×10-8 M 1.3×10-8 M 7.90


Acidification of the OceansDoes it matter? Yes, especially if you are a coral.

Shells and skeletons are made largely of CaCO3, which isslightly soluble:

CaCO3(s) W Ca2+ + CO32-

In seawater, Q > K, so organisms can extract CaCO3. IfCO3

2- decreases, this become more difficult.

We have [CO32-] = Ka1Ka2KeqPCO2 / [H3O

+]2

and [H3O+] . Ka1KeqPCO2 / [Alk]

so [CO32-] . Ka2[Alk]2 / (Ka1KeqPCO2)

Conclusion: More CO2 in the atmosphere means lower[CO3

2-] in the oceans. Possible problem for marine life.

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