acid base equilibria buffers and titrations

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

The presence of a common ion suppresses the ionization of a weak acid or a weak base.

Consider a mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).

CH3COONa (s) Na+ (aq) + CH3COO- (aq)

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

common ion

A buffer solution is a solution of:

1. A weak acid or a weak base and

2. The salt of the weak acid or weak base

Both must be present!

A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

Add strong acidH+ (aq) + CH3COO- (aq) CH3COOH (aq)

Add strong baseOH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)

Consider an equal molar mixture of CH3COOH and CH3COONa

HCl + CH3COO- CH3COOH + Cl-

Consider mixture of salt NaA and weak acid HA.

HA (aq) H+ (aq) + A- (aq)

NaA (s) Na+ (aq) + A- (aq)Ka =

[H+][A-][HA]

[H+] =Ka [HA]

[A-]

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

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

pH = pKa + log [A-][HA]

pKa = -log Ka

Henderson-Hasselbalch equation

pH = pKa + log [conjugate base][acid]

1. What is the pH of a solution containing 0.30 M HCOOH?(Ka of HCOOH = 1.7 x 10-4)

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.00

+x

x x

0.30 – x 0.30Ka = 1.7 x 10-4 = [H+][HCOO-] / [HCOOH]

= (x2) / (0.30)[H+] = 0.0071pH = -log[H+] = -log(0.0071)

= 2.14

2. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

Common ion effect0.30 – x 0.300.52 + x 0.52

pH = pKa + log [HCOO-][HCOOH]

HCOOH pKa = 3.77

pH = 3.77 + log [0.52][0.30]

= 4.01

Mixture of weak acid and conjugate base!

Buffer Preparation sample problem

What mass of sodium acetate must be dissolved in stock 0.250 Macetic acid to produce 300. mL of buffer solution with pH = 5.09?(Assume negligible volume change when salt is added to stock)

Equilibrium among the buffer components is expressed by:CH3COOH(aq) + H2O(l) ↔ H3O+(aq) + CH3COO–(aq)Ka = 1.8 x 10–5

pKa = −logKa = −log(1.8 x 10−5) = 4.74Henderson-Hasselbalch:

pH = pKa + log([CH3COO−] / [CH3COOH])5.09 = 4.74 + log([CH3COO−] / 0.250)0.35 = log([CH3COO−] / 0.250)2.24 = [CH3COO−] / 0.250[CH3COO−] = 0.56 M = [NaCH3COO]

So:mass NaCH3COO = (0.56 mol/L) (0.300 L) (82.0 g/mol) = 14 g NaCH3COO

Which of the following are buffer systems? (a)KF/HF, (b) KBr/HBr, (c) Na2CO3/NaHCO3

(d) Na2CO3/H2CO3

(a) HF is a weak acid and F- is its conjugate basebuffer solution

(b) HBr is a strong acidnot a buffer solution

(c) CO32- is a weak base and HCO3

- is its conjugate acidbuffer solution

(d) CO32- is a weak base but H2CO3 is NOT its conjugate acid

Buffer CapacityBuffer capacity is the ability to resist pH change.

The more concentrated the components of a buffer, the greaterthe buffer capacity.

The pH of a buffer is distinct from its buffer capacity.

A buffer has the highest capacity when the component concentrations are equal.

Buffer Range

Buffer range is the pH range over which the buffer acts effectively.

Buffers have a usable range within ± 1 pH unit of the pKa ofits acid component when [HA] and [A-] differ by ≤ 10.

Properly Choosing and Preparing a Buffer System

1. Choose the conjugate acid-base pair based on desired pH, keeping in mind that you want pH ≈ pKa when [HA] ≈ [A-].

2. Calculate the ratio of buffer component concentrations using the Henderson-Hasselbalch equation.

3. Determine the buffer concentration based on desired buffer capacity. The greater the [HA] and [A-], the greater the capacity of the buffer system. Usually 0.5 M is suitable for most lab uses.

4. Mix the solution and adjust the pH of the buffer system with small amounts of strong acid or base, using pH meter.

Buffer Preparation sample problem

How would you prepare a “phosphate buffer” with a pH of about 7.40?

For an effective buffer with optimal buffer capacity, pH should be ≈ pKa, which means that log([A–]/[HA]) ≈ 0, or [A–]/[HA] ≈ 1.

So, first choose the appropriate phosphate-containing acid:Option 1: H3PO4(aq) ↔ H+(aq) + H2PO4

–(aq) Ka = 7.5 x 10–3 pKa = 2.12Option 2: H2PO4

–(aq) ↔ H+(aq) + HPO42–(aq) Ka = 6.2 x 10–8 pKa = 7.21

Option 3: HPO42–(aq) ↔ H+(aq) + PO4

3–(aq) Ka = 4.8 x 10–11 pKa = 12.32

Option 2 is most suitable because its pKa is closest to the desired pH.

pH = pKa + log([A–]/[HA]) 7.40 = 7.21 + log([HPO4

2–]/[H2PO4–])

log([HPO42–]/[H2PO4

–]) = 0.19[HPO4

2–]/[H2PO4–] = 100.19 = 1.5

So, one way to make the buffer is to dissolve Na2HPO4 and NaH2PO4 in a mole ratio of 1.5:1.0 in water (e.g., 1.5 moles Na2HPO4 and 1.0 mole NaH2PO4 in enough water to make up a 1-liter solution).

Buffer Preparation via “Partial Neutralization”

Another way to prepare a buffer is to mix appropriate amounts of HA and strong base (e.g., NaOH) until the right amount of A- is produced.

For example:

To a certain HCOOH ionization equilibrium:

HCOOH(aq) + H2O(l) ↔ HCOO−(aq) + H3O+(aq)

add NaOH (OH-) which reacts with the HCOOH and produces HCOO- :

HCOOH(aq) + OH−(aq) → HCOO−(aq) + H2O(l) Keep adding the NaOH until [HCOOH] ≈ [HCOO-] or until the desired [HCOO-] / [HCOOH] ratio is achieved for the desired buffer pH.

If your resulting [HCOOH] and [HCOO-] are sufficiently high, and somewhat equal to each other, you’ve got a buffer system of good capacity.

= 9.20

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. The Ka of NH4

+ = 5.62 x 10-10. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?

NH4+(aq) + H2O (l) H3O+(aq) + NH3(aq)

pH = pKa + log [NH3][NH4

+]pKa = 9.25 pH = 9.25 + log [0.30]

[0.36]= 9.17

NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)

start (moles)

end (moles)

0.029 0.001 0.024

0.028 0.0 0.025

pH = 9.25 + log [0.25][0.28]

[NH4+] =

0.0280.100

final volume = 80.0 mL + 20.0 mL = 100 mL = 0.100 L

[NH3] = 0.0250.100

TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

The indicatorchanges color

(pink)

Strong Acid-Strong Base TitrationsNaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)

OH- (aq) + H3O+ (aq) 2H2O (l)

Weak Acid-Strong Base TitrationsCH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

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

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

At equivalence point (pH > 7) – why?

Weak Acid – Strong Base Titration sample problem

If it takes 12.155 mL of 0.1000 M NaOH titrant to reach the equivalence point during titration of a 25-mL sample of acetic acid solution (Ka of CH3COOH = 1.8 x 10−5 at 25°C),

a)what was the original molarity (M) of the acid solution before titration?

b)what was the pH of the original acid solution?

c)what was the pH after addition of 3.00 mL of titrant to the acid solution?

d)what was the pH when ½ of the original CH3COOH was neutralized?

e)what is the pH at the equivalence point?



a)what was the original molarity (M) of the acid solution before titration?

monoprotic acid and monoprotic base, so:

(moles of acid neutralized) = (moles of base required to neutralize)

[(volume of acid)(M of acid)] = [(volume of base)(M of base)]

[(0.025 L)(M of acid)] = [(0.01215 L)(0.1000 M base)

(M of acid) = 0.049 M



b) what was the pH of the original acid solution?

CH3COOH(aq) + H2O(l) ↔ CH3COO−(aq) + H3O+(aq) Ka = 1.8 x 10−5

Ka = 1.8 x 10−5 = [H3O+][CH3COO−] / [CH3COOH] = x2 / 0.049 M

x = [H3O+] = 9.4 x 10−4 so pH = −log (9.4 x 10−4) = 3.03

(or, pH = −log (Ka • [CH3COOH])1/2 = −log [(1.8x10−5)(0.049)]1/2 = 3.03



c) what was the pH after addition of 3.00 mL of titrant to the acid solution?

CH3COOH(aq) + OH−(aq) → CH3COO−(aq) + H2O(l)I: (vol. x M) (vol. x M) 0

(0.025 L x 0.049 M) = (0.00300 L x 0.1000 M) =

0.0012 mol 0.000300 mol (limiting reactant)

C: −0.000300 mol −0.000300 mol +0.000300 molEnd: 0.0009 mol 0 mol 0.0003 mol

Buffer solution produced: pH = pKa + log [(CH3COO−) / (CH3COOH)] pH = 4.74 + log [(0.0003) / (0.0009)] pH = 4.74 + log [1/3] = 4.26



d)what was the pH when ½ of the original CH3COOH was neutralized?

When ½ of the acid is neutralized, that means that you have a buffer solution where [CH3COOH]eq = [CH3COO−]eq :

pH = pKa + log [(CH3COO−) / (CH3COOH)] pH = 4.74 + log 1 pH = 4.74 + 0 = 4.74

(this occurs in the midway point of the “buffer zone”, where pH = pKa)



e)what is the pH at the equivalence point?

At the eq. pt., the titration is over and the original acid has been completely neutralized:

CH3COOH(aq) + OH−(aq) → CH3COO−(aq) + H2O(l)I: (vol. x M) (vol. x M) 0

(0.025 L x 0.049 M) = (0.01215 L x 0.1000 M) ≈

0.0012 mol 0.0012 molC: −0.0012 mol −0.0012 mol +0.001200 molEnd: 0 mol 0 mol 0.0012 mol

So, [CH3COO−] at the eq. pt. = [(0.0012 mol) / (0.025 L + 0.012 L)] = 0.032 M

Now the CH3COO− hydrolyzes water in accordance to its Kb :CH3COO−(aq) + H2O(l) ↔ CH3COOH(aq) + OH−(aq)

Kb = 1x10−14 / 1.8x10−5 = 5.6x10−10 = x2 / 0.032 so x = [OH−] = 4.2x10−6 MpOH = −log(4.2x10−6) = 5.37, so pH = 14.00 − 5.37 = 8.63

Strong Acid-Weak Base TitrationsHCl (aq) + NH3 (aq) NH4Cl (aq)

NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq)

At equivalence point (pH < 7):

H+ (aq) + NH3 (aq) NH4Cl (aq)

Which indicator(s) would you use for a titration of HNO2 with NaOH ?

Weak acid titrated with strong base.At equivalence point, will have conjugate base of weak acid.At equivalence point, pH > 7Use cresol red or phenolphthalein

acid base equilibria buffers and titrations

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