Question

A chemist titrates 230.0mL of a 0.4292M hydrocyanic acid HCN solution with 0.6266M KOH solution at 25°C. 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 KOH solution added.

Answer 1

pH at equivalence = 11.31

Explanation

moles HCN = (concentration HCN) * (volume HCN)

moles HCN = (0.4292 M) * (230.0 mL)

moles HCN = 98.716 mmol

moles KOH required = moles HCN

moles KOH required = 98.716 mmol

volume KOH required = (moles KOH required) / (concentration KOH)

volume KOH required = (98.716 mmol) / (0.6266 M)

volume KOH required = 157.5 mL

Total volume at equivalence point = (volume HCN) + (volume KOH required)

Total volume at equivalence point = (230.0 mL) + (157.5 mL)

Total volume at equivalence point = 387.5 mL

At equivalence point, all HCN is neutralized to CN^-

concentration CN^- at equivalence = (moles HCN) / (Total volume at equivalence point)

concentration CN^- at equivalence = (98.716 mmol) / (387.5 mL)

concentration CN^- at equivalence = 0.2547 M

pK_a = 9.21

pK_b = 14 - pK_a

pK_b = 14 - 9.21

pK_b = 4.79

K_b = 10^-pK_b

K_b = 10^-4.79

K_b = 1.62 x 10^-5

ICE table	CN- (aq)	H2O (l)		HCN (aq)	OH- (aq)
Initial conc.	0.2547 M	-		0	0
Change	-x	-		+x	+x
Equilibrium conc.	0.2547 M - x	-		+x	+x

K_b = [HCN]_eq[OH^-]_eq / [CN^-]_eq

1.62 x 10^-5 = [(x) * (x)] / (0.2547 M - x)

Solving for x, x = 2.02 x 10^-3 M

[OH^-] = x = 2.02 x 10^-3 M

pOH = -log[OH^-]

pOH = -log(2.02 x 10^-3 M)

pOH = 2.69

pH = 14 - pOH

pH = 14 - 2.69

pH = 11.31