pH and pKa Henderson-Hasselbalch Approach

Question

Calculate the pH of a buffer solution that is 0.050 M in benzoic acid (HC7H5O2) and

0.150 M in sodium benzoate (NaC7H5O2).

For benzoic acid, Ka = 6.5 × 10–5.

Please show with ICE table too.

Answer 1

The Henderson Hasselbalch equation for a weak acid buffer is...

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

In the current problem, we are asked to find the pH. We are given the Ka, so we can find the pKa. We are also given the [conjugate base] and [acid]. Thus, we have everything we need to find the pH. You asked to show with ICE table, but since there is no addition of an acid or a base, we don't really have the need for an ICE table. This is useful when asked to find pH AFTER the addition of either an acid or a base.

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

pKa = -log Ka = -log 6.5x10^-5 = 4.19

[conjugate base] = 0.150 M NaC₇H₅O₂

[acid] = 0.050 M HC₇H₅O₂

Solving for pH...

pH = 4.19 + log (0.150/0.050)

pH = 4.19 + log 3

pH = 4.19 + 0.48

pH = 4.67

Edit:

The equation for dissociation of the acid, in case you wanted to use it in an ICE table is...

HC₇H₅O₂ + H₂O <===> C₇H₅O₂^- + H₃O⁺