Question

An analytical chemist is titrating 167.8 mL of a 1.000 M solution of cyanic acid (HCNO) with a 0.4700 M solution of NaOH. The p Ka of cyanic acid is 3.46. Calculate the pH of the acid solution after the chemist has added 236.2 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.

Answer 1

Given pKa = 3.46

The initial moles of HCNO = 1.000 M x 167.8 ml

= 167.8 mmol

Moles of NaOH added = 0.4700 M x 236.2 ml

= 111.014 mmol

The moles of NaCNO formed = 111.014 mmol

The moles of HCNO remained = 167.8 - 111.014

= 56.786 mmol

Let's consider Henderson - Hasselbach equation ,

pH = pKa + log(base/acid)

= 3.46 + log (111.014 / 56.786 )

= 3.46 + 0.290

pH = 3.75