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A 0.495 M solution of nitrous acid, HNO2, has a pH of 1.83.b. Write the equilibrium...
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The first part of this question is answered at;
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`
Posted by jeew-m on July 3, 2013 at 12:25 PM (Answer #1)
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