Homework Help: Chemistry

Posted by Tiri on Friday, April 6, 2012 at 4:57pm.

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?

Chemistry - DrBob222, Friday, April 6, 2012 at 5:21pm
..........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.

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