Question

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250?

Answer 1

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons
`\text{H}^(+)` than hydroxide ions
`\text{OH}^(-)`. The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that
`x \; \text{L}` of the 0.307
`\text{mol} \cdot \text{dm}^(-3)` sodium hydroxide solution was added to the acetic acid. Based on previous reasoning,
`x` is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain
`0.2000 * 0.425 - 0.307 \; x = 0.085 - 0.307 \; x` moles of acetic acid and
`0.307 \; x` moles of acetate ions.

Let
`\text{HAc}` denotes an acetic acid molecule and
`\text{Ac}^(-)` denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

`\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^(+) & + & \text{Ac}^(-)\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}`

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of
`10^(4.250)\; \text{mol}\cdot \text{dm}^(-3)`. There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of
`10^(-4.250)\;\text{mol}\cdot \text{dm}^(-3)` in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by
`10^(-4.250)\;\text{mol}\cdot \text{dm}^(-3)` would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

By definition:

`\text{K}_(a) = [\text{H}^(+)] \cdot [\text{Ac}^(-)] / [\text{HAc}]\\\phantom{\text{K}_(a)} = 10^(-4.250) * (10^(-4.250) + 0.307 \; x) / (0.085 - 10^(-4.250) - 0.307 \; x)`

The question states that

`\text{K}_(a) = 1.75 * 10^(-5)`

such that

`10^(-4.250) * (10^(-4.250) + 0.307 \; x) / (0.085 - 10^(-4.250) - 0.307 \; x) = 1.75 * 10^(-5)\\6.16 * 10^(-5) \; x = 1.48 * 10^(-6)\\x = 0.0241`

Thus it takes
`0.0241 \; \text{L}` of sodium hydroxide to produce this buffer solution.