The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Part A Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 6.3×10−2 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 6.3×10−2 M . Part C Calculate the equilibrium concentration of C6H5COOH in the solution if the initial concentration of C6H5COOH is 6.3×10−2 M . Express your answer using two significant figures.
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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