# A 0.495 M solution of nitrous acid, HNO2, has a pH of 1.83.b. Write the equilibrium expression and calculate the value of Ka for nitrous acid.

The first part of this question is answered at;

http://www.enotes.com/homework-help/0-495-m-solution-nitrous-acid-hno2-has-ph-1-83-442016

There we got the `[H^+] = 0.0148M`

`HNO_2 harr H^++NO_2^-`

So we can say that at equilibrium state;

`[H^+] = [NO_2^-] = 0.0148M`

`[HNO_2] = 0.495-2xx0.0148 = 0.465M`

`K_a = ([H^+][NO_2^-])/[HNO_2]`

`K_a = (0.0148)^2/(0.465)`

`K_a =...

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The first part of this question is answered at;

http://www.enotes.com/homework-help/0-495-m-solution-nitrous-acid-hno2-has-ph-1-83-442016

There we got the `[H^+] = 0.0148M`

`HNO_2 harr H^++NO_2^-`

So we can say that at equilibrium state;

`[H^+] = [NO_2^-] = 0.0148M`

`[HNO_2] = 0.495-2xx0.0148 = 0.465M`

`K_a = ([H^+][NO_2^-])/[HNO_2]`

`K_a = (0.0148)^2/(0.465)`

`K_a = 4.71xx10^(-4)M `

So the `K_a` for the reaction is `4.71xx10^(-4)M`