Question

A 0.150 M solution of nitrous acid (HNO2) is made. Ka = 4.5 x 10-4

1. Show the equilibrium which occurs when this acid is dissolved in water.
2. What is the pH of the solution? Show all work clearly.
3. 100.0 mL of the solution is titrated against 0.150 M NaOH. What is the pH at the equivalence point?
4. What is true about the pH and pKa, and the [acid] and [conjugate base] at the half-equivalence point?

Answer 1

..........HNO2 + H2O ==> H3O^+ + NO2^-
initial...0.150M..........0........0
change.....-x.............x........x
equil...0.150-x...........x.........x

Ka = (H3O^+)(NO2^-)/(HNO2)
Substitute from the ICE chart and solve for x = (H3O^+). Then pH = -log(H3O^+)
3.
Calculate volume NaOH needed to arrive at the equivalence point and from that the the concn of the salt at that point. The pH at the equivalence point is determined by the hydrolysis of the salt. I will call the concn of the salt C.
.........NO2^- + HOH ==> HNO2 + OH^-
initial...c................0.....0
change...-x...............x.......x
equil....c-x...............x......x

Kb for NO2^- = (Kw/Ka for HNO2) = (HNO2)(OH^-)/(HNO2)
Substitute and solve for x = (OH^-), then convert to pH.

At the half way point pH = pKa.