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vovangra [49]
3 years ago
12

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.)
Chemistry
1 answer:
Fynjy0 [20]3 years ago
8 0

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}

<u>Balance: </u>

[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

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