Question

I've seen similar problems, but I still do not get it! Can someone show me step by step?
A 34.8 mL sample of a 0.588 M aqueous nitrous acid solution is titrated with a 0.220 M aqueous sodium hydroxide solution. What is the pH after 29.9 mL of base have been added?

Ka of Nitrous Acid: 3.39

Answer 1

Given:
pKa = 3.39


Given:
M(HNO2) = 0.588 M
V(HNO2) = 34.8 mL
M(NaOH) = 0.22 M
V(NaOH) = 29.9 mL


mol(HNO2) = M(HNO2) * V(HNO2)
mol(HNO2) = 0.588 M * 34.8 mL = 20.4624 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.22 M * 29.9 mL = 6.578 mmol


We have:
mol(HNO2) = 20.4624 mmol
mol(NaOH) = 6.578 mmol

6.578 mmol of both will react

excess HNO2 remaining = 13.8844 mmol
Volume of Solution = 34.8 + 29.9 = 64.7 mL
[HNO2] = 13.8844 mmol/64.7 mL = 0.2146M

[NO2-] = 6.578/64.7 = 0.1017M

They form acidic buffer
acid is HNO2
conjugate base is NO2-


use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.39+ log {0.1017/0.2146}
= 3.066


Answer: 3.07