Question

1. A 41.6 mL sample of a 0.466 M aqueous hydrocyanic acid solution is titrated with a 0.445 M aqueous solution of potassium hydroxide. How many milliliters of potassium hydroxide must be added to reach a pH of 9.215?

_______ mL

Answer 1

HCN <----> H+ + CN-

Ka = 6.17 x 10-10
pKa = 9.21

We have 41.6 mL of 0.466 M HCN titrated with 0.455 M of KOH. We need to find how many mL of KOH need to reach pH = 9.215.
pKa = pH of the solution;

Stoichiometry:

HCN + KOH ----> H2O + KCN

KOH is a strong base. it will react with the HCN until all of the H+ is neutralized

At midpoint, pH = pKa
moles of HCN = (0.0416 L)(0.466 M) = 0.02 mol
You could prove this by finding that it would take 0.02 mol OH- added to reach equivalence, divide by 2 to get the midpoint,
moles of OH- = (Volume L)(0.455 M) = 0.2 mol
volume L = 0.02/ 0.455 = 0.044 L
to get the midpoint = volume/2 = 0.044/2 = 0.022 L
volume required = 22 mL