Calculate the percent dissociation of 0.40 M benzoic acid, C6H5COOH. (Ka = 6.3 x 10^-5) ___%?

Calculate the percent dissociation of 0.40 M benzoic acid, C6H5COOH. (Ka = 6.3 x 10^-5)

___%?

Expert Solution

Dear friend your answer is

% Dissociation

C6H5COOH(aq) + H2O(l) <===> C6H5COO-(aq) + H3O+(aq)

I (M) .40 --- 0 0

C(M) -x --- +x +x

E(M) .40-x --- x x

K_a = 6.3 x 10 ^{– 5}= (x)(x)/(0.40 -x)

Approximation: 0.40 -x ~ 0.40

so x = 0.0050199 M

(x/0.40)*100% = (0.0050199/.40)*100% = 1.2549 %

Percent dissociation 1.2549 %

Thank you

queen_honey_blossom answered 2 years ago

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Calculate the percent dissociation of 0.40 M benzoic acid, C6H5COOH. (Ka = 6.3 x 10^-5) ___%?

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