Question

Suppose you wanted to produce an aqueous solution of pH = 8.80 by dissolving KNO2 in water, at what molarity?

Here is my work:

KNO2 ----> K^+ NO2-

NO2^- + H2O ---> HNO2 + OH^-

Ka= 4.5 x 10^-4


Ka*Kb=Kw

Kb=Kw/Ka

kb=10^-14/4.5 x 10^-4=2.222 x 10^-11

pH+pOH=14.00

pOH=14.00-8.80=5.20

[OH^-]=10^-(5.20)= 6.310 x 10^-6 M

OH=HNO2

Kb=[6.310 x 10^-6 M][6.310 x 10^-6 M]/[x-6.31 x 10^-6 M]

Kb*[x-6.31 x 10^-6 M]=3.981 x 10^-11 M

(3.981 x 10^-11)/(2.222 x 10^-11)=x-6.31 x 10^-6 M

1.8 M=x

x=KNO2

Answer 1

I obtained 1.79 M which I would round to 1.8M; however, you made a typo on line 11 counting down from the KNO2 equation line. You didn't square the 6.31E-6 in the numerator but you corrected that in the next line and you have the right number there of 3.98E-11.

Answer 2

Thought is correct, but just wanted to make sure. Thanks for catching the typo.

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