Question

Q2/ Calculate the degree of dissociation and pH of a 0.15 moldm-3 solution of weak acid for which Ka = 3×10-5 ?

Answer 1

Consider the acid's dissociation equilibrium in solution,

`HA rightleftharpoons H^(+) + A^(-)`

where,

`K_"a" = ([H^+][A^-])/([HA]) approx 3*10^-5`

Now, let's construct an ICE table for the same equilibrium,

`HA rightleftharpoons H^(+) + A^(-)`

Hence,

`K_"a" = x^2/(0.15 -x) = 3*10^-5`

`=> x = [H^+] approx 2.11*10^-3`

and by extension,

`"pH" = -log[H^+] approx 2.68`