Question

2.Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M KCN with 4.33 mL of 0.425 M strong acid HCl. HCN pKa = 9.21.

Answer 1

Given:
pKa = 9.21

use:
pKa = -log Ka
9.21 = -log Ka
Ka = 6.166*10^-10
Given:
Ka = 6.166*10^-10

use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/6.166*10^-10
Kb = 1.622*10^-5

Given:
M(HCl) = 0.425 M
V(HCl) = 4.33 mL
M(CN-) = 0.1 M
V(CN-) = 50 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.425 M * 4.33 mL = 1.8402 mmol

mol(CN-) = M(CN-) * V(CN-)
mol(CN-) = 0.1 M * 50 mL = 5 mmol



We have:
mol(HCl) = 1.8402 mmol
mol(CN-) = 5 mmol

1.8402 mmol of both will react
excess CN- remaining = 3.1597 mmol
Volume of Solution = 4.33 + 50 = 54.33 mL
[CN-] = 3.1597 mmol/54.33 mL = 0.0582 M
[HCN] = 1.8402 mmol/54.33 mL = 0.0339 M

They form basic buffer
base is CN-
conjugate acid is HCN

Kb = 1.622*10^-5

pKb = - log (Kb)
= - log(1.622*10^-5)
= 4.79

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.79+ log {3.387*10^-2/5.816*10^-2}
= 4.555

use:
PH = 14 - pOH
= 14 - 4.5552
= 9.4448

Answer: 9.44