Question

A weak acid (HA) has a pKa of 4.011. If a solution of this acid has a pH of 4.141, what percentage of the acid is not ionized? (Assume all H in solution came from the ionization of HA.)

Answer 1

Given, pKa (weak acid HA)= 4.679

pH= 4.876

Now, Dissociation equation for weak acid HA:

HA ----> H+ + A-

I M 0 0

C -x +x +x

E (M-x) +x +x

Thus, ka = [H+][A-]/[HA]

ka= (x* X)/(M-x)

Now, Calculating [H+] and ka:

We know that, pH= -log[H+]

So, [H+]= 10^-pH

= 10^-4.141 = 7.22*10^-5

Also, pKa= -log Ka

Ka= 10^-pka= 10^-4.011= 9.75*10^-5

Substituting given in Ka equation,

We have, 9.75*10^-5 = (7.22*10^-5)(7.22*10^-5)/(M- 7.22*10^-5).

(M-x)= 5.34*10^-5

M= (5.34+7.22)*10^-5

M= 1.256*10^-4

non ionised ion= Total concentration - ionised concentration.

Non ionised ion= 12.56*10^-5 - 7.22*10^-5

= 5.34*10^-5.

Percentage= (Non ionised ion/Total ion)*100

=(5.34*10^-5/12.56*10^-5)*100

= 42.51%