$$pH = pK_a + log \frac{[A^-]}{[HA]}$$
where:
* pH is the acidity of the solution
* pKa is the acid dissociation constant of the weak acid
* [A-] is the concentration of the conjugate base of the weak acid
* [HA] is the concentration of the weak acid
We can use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution. We know the pH of the buffer solution we want to make (4.75) and the pKa of benzoic acid (4.20). We can also assume that the concentration of the conjugate base of benzoic acid is equal to the concentration of sodium benzoate. Therefore, we can rearrange the Henderson-Hasselbalch equation to solve for the concentration of benzoic acid:
$$[HA] = \frac{[A^-]}{10^{pH - pK_a}}$$
Substituting the values we know into the equation, we get:
$$[HA] = \frac{0.025}{10^{4.75 - 4.20}} = 0.0040 M$$
The total concentration of benzoic acid and sodium benzoate in the buffer solution is 0.029 M. Therefore, the volume of 0.200 M benzoic acid we need to add to make 500 mL of the buffer solution is:
$$V_{BA} = \frac{(0.0040 M)(500 mL)}{0.200 M} = 10.0 mL$$
The volume of 2.00 M sodium hydroxide we need to add to make 500 mL of the buffer solution is:
$$V_{NaOH} = \frac{(0.025 M)(500 mL)}{2.00 M} = 6.25 mL$$
Therefore, we need to add 10.0 mL of 0.200 M benzoic acid and 6.25 mL of 2.00 M sodium hydroxide to make 500 mL of a buffer solution with the same pH as one made from 475 benzoic acid 25 naoh.
