Question

An analytical Chemistis tiltrating 96.8 ml of a 1.200 M solution of cyanic acid (HCNO) with a 1.200 M solution of NaOH. The pk of cyanic add is 3.46. Calculate the pH of the acid solution after the chemist has added 105. mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places pH-1

Answer 1

use:

pKa = -log Ka

3.46 = -log Ka

Ka = 3.467*10^-4

Given:

M(HCNO) = 1.2 M

V(HCNO) = 96.8 mL

M(NaOH) = 1.2 M

V(NaOH) = 105 mL

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

mol(HCNO) = 1.2 M * 96.8 mL = 116.16 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 1.2 M * 105 mL = 126 mmol

We have:

mol(HCNO) = 116.16 mmol

mol(NaOH) = 126 mmol

116.16 mmol of both will react

excess NaOH remaining = 9.84 mmol

Volume of Solution = 96.8 + 105 = 201.8 mL

[OH-] = 9.84 mmol/201.8 mL = 0.0488 M

use:

pOH = -log [OH-]

= -log (4.876*10^-2)

= 1.3119

use:

PH = 14 - pOH

= 14 - 1.3119

= 12.6881

Answer: 12.69