Question

At a given temperature the pH of 0.25M solution of a weak acid is 3.8. What is the value of Ka for this acid?

Answer is 1x10^-7 but I don't understand how. Please show work

Answer 1

Since it has single K_a value, hence the acid must be a monobasic weak acid. let the monobasic weak acid be HA, where H denotes hydrogen and A denotes the anion.

Given concentration of the solution, C = 0.25M

P^H  = 3.8

since it is a weak acid, it is partially dissociated. Let the degree of dissociation be "x." Now the dissociation reaction can be written as

HA ------------> H⁺   + A^-

C(1-x) Cx Cx

K_a for the above dissociation ofthe above acid can be calculated as

K_a = ([H⁺][A^-])/[HA] = (Cx)(Cx)/C(1-x) =Cx²/(1-x)

since x <<<1, hence (1-x) nearly equals to 1. Hence

K_a = Cx²  

=>x = (K_a/C)^1/2

Hence the concentration of H⁺ = Cx = C(K_a/C)^1/2 = (K_aC)^1/2

Now the P^H of the solution can be calculated as

P^H = -log[H⁺] = -log(K_aC)^1/2 = -1/2[logK_a + logC] = 3.8

=> logK_a  = - 7.6 - log0.25 = -7.6 - (-0.6020) = -7.6+0.6020 = -7

=> K_a  = antilog(-7) = 1x10^-7  (answer)