Start by setting up an ICE table:
We have the following reaction:
HA(aq)+H_2O(aq) rightleftharpoons A^(-)(aq) + H_3O^(+)(aq)HA(aq)+H2O(aq)⇌A−(aq)+H3O+(aq)
And we have an initial concentration of HAHA at 0.64 moldm^-3moldm−3, so let's plug what we have into the ICE table:
color(white)(mmmmmi)HA(aq) + H_2O(l) rightleftharpoons A^(-)(aq)+ H_3O^(+)(aq)mmmmmiHA(aq)+H2O(l)⇌A−(aq)+H3O+(aq)
"Initial:"color(white)(mm)0.64color(white)(miimm)-color(white)(mmmmm)0color(white)(mmmmmm)0Initial:mm0.64miimm−mmmmm0mmmmmm0
"Change:"color(white)(im)-xcolor(white)(miimm)-color(white)(mmmm)+xcolor(white)(mmmmii)+xChange:im−xmiimm−mmmm+xmmmmii+x
"Eq:"color(white)(mmm)0.64-xcolor(white)(iimm)-color(white)(mmmmm)xcolor(white)(mmmmmm)xEq:mmm0.64−xiimm−mmmmmxmmmmmmx
Now using the K_aKa expression:
K_a=([H_3O^(+)] times [A^(-)])/[[HA]]Ka=[H3O+]×[A−][HA]
From our ice table and the values given, we can plug all of the equilibrium values into the K_aKa expression as K_aKa is constant.
(6.3 times10^-5)=(x^2)/(0.64-x)(6.3×10−5)=x20.64−x
However, the change in concentration of the acid can be considered negligible, due to K_aKa being small: (0.64-x=0.64)(0.64−x=0.64)
The equation above can also be solved by setting up a quadratic equation, but you save time by making the assumption that the change in concentration is negligible - and it rounds off to the same answer.
(6.3 times10^-5)=(x^2)/(0.64)(6.3×10−5)=x20.64
Hence:
x=0.0063498031x=0.0063498031
There the equation becomes:
[H_3O^(+)]=x=0.0063498031[H3O+]=x=0.0063498031
pH=-log[H_3O^(+)]pH=−log[H3O+]
pH=-log[0.0063498031]pH=−log[0.0063498031]
pH approx 2.2pH≈2.2
