Question

An analytical chemist is titrating 109.7mL of a 0.2200M solution of hydrazoic acid HN3 with a 0.3700M solution of NaOH. The pKa of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 43.91mL of the NaOH solution to it. Round your answer to 2 decimal places

Answer 1

The initial moles of acid and those of NaOH added are calculated:

n Acid = M * V = 0.22 M * 0.1097 L = 0.024 mol

n NaOH = 0.37 M * 0.04391 L = 0.016 mol

The NaOH reacts with the acid and forms salt, the pH is calculated:

pH = pKa + log (n Salt / n Acid) = 4.72 + log (0.016 / 0.024 - 0.016) = 5.02

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