Question

What concentrations of acetic acid (pKa=4.76)(pKa=4.76) and acetate would be required to prepare a 0.10 M0.10 M buffer solution at pH 4.6pH 4.6? Note that the concentration, pH value, or both may differ from that in the first question.

Strategy

1. Rearrange the Henderson–Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A−]/[HA][A−]/[HA] .

2. Use the mole fraction of acetate to calculate the concentration of acetate.

3. Calculate the concentration of acetic acid.

Step 1: The ratio of base to acid is [acetate][acetic acid]=0.69[acetate][acetic acid]=0.69. That is, there are 0.69 molecules0.69 molecules of acetate for each molecule of acetic acid.

Step 2: Use the mole fraction of acetate to calculate the concentration of acetate.

Answer 1

Buffer Concentration = 0.10 M

Target pH = 4.6

pKa = 4.76

Hence, using the Henderson-Hasselbalch equation, we can find the ratio of each component of the buffer

pH = pka + log Р [НА)

Where

A-] = acetate

H4 = acetic acid

Putting in the values,

1 4-] pH = pka + log - SHA [A] 4.6 = 4.76 + log » [НА) 101 [A] = 4.6 – 4.76 = -0.16 [НА = 147 = 10-0.16 – 0.69 = 0,09 =[A-]=0.69[H A...... (1) 1

Our target buffer concentration = 0.10 M

Hence,

A-]+HA= 0.10 M.... (2)

Solving the two linear equations (1) and (2) for the two unknown variable,

[A-] + H A = 0.10 M 0.69[HA] + [HA] = 0.10 M →1.69[HA] = 0.10 M 0.10 [HA] = [acetic acid) = 16M 0.059 M

Again, substituting the value of [HA] in equation (2),

[A] + [HA] = 0.10 M [A-] = 0.10 M - [HA] = 0.10 M -0.059 M = 0.041 M (acetate) = 0.041 M

Hence, the concentrations of acetic acid and acetate required for the target buffer are 0.059 M and 0.041 M respectively.