Question

A chemist titrates 210.0 mL of a 0.7066 M hydrocyanic acid (HCN) solution with 0.4210 M NaOH solution at 25C. Calculate the pH at equivalence. The pKa of hydrocyanic acid is 9.21.

Round your answer to 2 decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added.

Answer 1

The balanced equation for the reaction is:

HCN (aq) + NaOH (aq) → NaCN (aq) + H2O (l)

First, we need to find the volume of NaOH solution needed to reach equivalence:

0.7066 M HCN × 0.2100 L = x M NaOH × 0.2100 L

x = 0.3366 M NaOH

The volume of NaOH solution needed to reach equivalence is:

0.3366 M NaOH × VNaOH = 0.4210 M NaOH × 0.2100 L

VNaOH = 0.527 L = 527 mL

So, the total volume of the solution at equivalence is:

210.0 mL + 527 mL = 737 mL = 0.737 L

Now we can use the Henderson-Hasselbalch equation to find the pH at equivalence:

pH = pKa + log([A-]/[HA])

At equivalence, [HCN] = [CN-], so:

pH = pKa + log(1) = pKa = 9.21

Therefore, the pH at equivalence is 9.21.