Question

An analytical chemist is titrating 201.8 mL of a 0.5300 M solution of nitrous acid (HNO, with a 1.100 M solution of KOH. The pk of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 116.5 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. D# = D x 5 ?

Answer 1

use:

pKa = -log Ka

3.35 = -log Ka

Ka = 4.467*10^-4

Given:

M(HNO2) = 0.53 M

V(HNO2) = 201.8 mL

M(KOH) = 1.1 M

V(KOH) = 116.5 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.53 M * 201.8 mL = 106.954 mmol

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

mol(KOH) = 1.1 M * 116.5 mL = 128.15 mmol

We have:

mol(HNO2) = 106.954 mmol

mol(KOH) = 128.15 mmol

106.954 mmol of both will react

excess KOH remaining = 21.196 mmol

Volume of Solution = 201.8 + 116.5 = 318.3 mL

[OH-] = 21.196 mmol/318.3 mL = 0.0666 M

use:

pOH = -log [OH-]

= -log (6.659*10^-2)

= 1.1766

use:

PH = 14 - pOH

= 14 - 1.1766

= 12.8234

Answer: 12.82