Question

Determine the pH at the equivalence point for the titration of 40.0 mL of 0.1 M HNO2 by 0.200 M KOH (Example 16.7). (The pKa of nitrous acid is 3.36.)

Answer 1

Answer:

$pH=13.28$

Explanation:

Nitrous acid

$HNO_2 \longrightarrow H^+ + NO_2^-$

$Ka=\frac{[H^+][NO_2^-]}{[HNO_2]}$

$pKa=-log(Ka)=3.36$

$Ka=4.36*10^{-4}$

Balance:

$[H^+]=X$

$[NO_2^-]=X$

$[HNO_2]=0.1-X$

$Ka=\frac{X*X}{0.1-X}$

$4.36*10^{-4}=\frac{X^2}{0.1-X}$

$4.36*10^{-5}-4.36*10^{-4}*X=X^2$

Solving:

$X=6.4*10^{-3}$

The NaOH is in excess:

$[NaOH]_f=0.2-0.0064=0.1936$

$pH=14+log([OH^-])$

$pH=14+log(0.1936)=13.28$

Answer 2

Answer:

pH = 8.18

Explanation:

pKa=-log(Ka)=3.36

Ka=4.36*10^-4

Kw=Ka*Kb

Kb=Kw/Ka

__________________________

Use ICE table:

HNO2- =.100-x

H+ = x

NO2- = x

Kb = x^2/.100-x (x is small) -> Kb = x^2/.100

x = 1.514*10^-6 M OH-

______________________

pOH=-log[OH-] = 5.819

pH=14-5.819 = 8.18