A 500.0-mL buffer solution is 0.10 M in benzoic acid and 0.10 M in sodium benzoate and has an initial pH of 4.19. What is the pH of the buffer upon addition of 0.010 mol of NaOH?

pH after the addition of NaOH is 4.37

Explanation:

When the amount of the weak acid = Amount of the conjugate base (As in the problem) pH = pKa. That means pKa of benzoic buffer is 4.19.

Now, to solve the pH of the buffer we need to use H-H equation for benzoic buffer:

pH = pKa + log [Benzoate] / [Benzoic acid]

pH = 4.19+ log [Benzoate] / [Benzoic acid]

You can take [] concentrations as the moles of both species.

When you add NaOH to the buffer, it reacts with benzoic acid producing more sodium benzoate and water, thus:

NaOH + Benzoic Acid → Sodium benzoate + Water.

Before the reaction, moles of benzoic acid and sodium benzoate were:

500.0mL = 0.500L × (0.10mol / L) = 0.050 moles

After the reaction, 0.010 moles of Benzoic acid are consumed and the same 0.010 moles of sodium benzoate are produced. That means moles of both species after reaction are:

[Benzoate] = 0.050 moles + 0.010 moles = 0.060 moles

[Benzoic acid] = 0.050 moles - 0.010 moles = 0.040 moles

Replacing in H-H equation:

pH = 4.19+ log [Benzoate] / [Benzoic acid]

pH = 4.19+ log [0.060mol] / [0.040mol]

pH = 4.37

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