Question

An analytical chemist is titrating 158.3 mL of a 0.5300 M solution of cyanic acid (HCNO) with a 1.100 M solution of KOH. The pk of cyanic acid is 3.46. Calculate the pH of the acid solution after the chemist has added 84.38 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places. | PH = 0 1 x 5 ? |

Answer 1

use:

pKa = -log Ka

3.46 = -log Ka

Ka = 3.467*10^-4

Given:

M(HCNO) = 0.53 M

V(HCNO) = 158.3 mL

M(KOH) = 1.1 M

V(KOH) = 84.38 mL

mol(HCNO) = M(HCNO) * V(HCNO)

mol(HCNO) = 0.53 M * 158.3 mL = 83.899 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 1.1 M * 84.38 mL = 92.818 mmol

We have:

mol(HCNO) = 83.899 mmol

mol(KOH) = 92.818 mmol

83.899 mmol of both will react

excess KOH remaining = 8.919 mmol

Volume of Solution = 158.3 + 84.38 = 242.68 mL

[OH-] = 8.919 mmol/242.68 mL = 0.0368 M

use:

pOH = -log [OH-]

= -log (3.675*10^-2)

= 1.4347

use:

PH = 14 - pOH

= 14 - 1.4347

= 12.5653

Answer: 12.57