The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×105. Part A Ca

Question

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×105. Part A Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 6.3×102 M . Express your answer using two significant figures. Part B Calculate the equilibrium concentration of C6H5COO in the solution if the initial concentration of C6H5COOH is 6.3×102 M . Part C Calculate the equilibrium concentration of C6H5COOH in the solution if the initial concentration of C6H5COOH is 6.3×102 M . Express your answer using two significant figures.

Explanation / Answer

Part A : C6H5COOH + H2O <==> C6H5COO- + H3O+

Ka = [C6H5COO-][H3O+]/[C6H5COOH]

let x amount of acid has dissociated at equilibrium

6.3 x 10^-5 = x^2/(6.3 x 10^-2 - x)

3.97 x 10^-6 - 6.3 x 10^-5x = x^2

x^2 + 6.3 x 10^-5x - 3.97 x 10^-6 = 0

Equilibrium concentration x = [H3O+] = 1.96 x 10^-3 M

Part B : Again repeating same procedure as above with [C6H5COOH] initial = 6.3 x 10^-2,

6.3 x 10^-5 = x^2/(6.3 x 10^-2 - x)

equilibrium concentration of [C6H5COO-] = x = 1.96 x 10^-3 M

Part C : Equilibrium concentration of [C6H5COOH] = 6.3 x 10^-2 - 1.96 x 10^3 = 6.10 x 10^-2 M

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