Question

What's the equilibrium pH of an initially 0.64M solution of the monoprotic acid benzoic acid (HA) at `25^@ "C"` (Ka = 6.3 x 10^-5) ?

Answer 1

See below:

Explanation:

Start by setting up an ICE table:

We have the following reaction:

`HA(aq)+H_2O(aq) rightleftharpoons A^(-)(aq) + H_3O^(+)(aq)`

And we have an initial concentration of `HA` at 0.64 `moldm^-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)`

`"Initial:"color(white)(mm)0.64color(white)(miimm)-color(white)(mmmmm)0color(white)(mmmmmm)0`

`"Change:"color(white)(im)-xcolor(white)(miimm)-color(white)(mmmm)+xcolor(white)(mmmmii)+x`

`"Eq:"color(white)(mmm)0.64-xcolor(white)(iimm)-color(white)(mmmmm)xcolor(white)(mmmmmm)x`

Now using the `K_a` expression:

`K_a=([H_3O^(+)] times [A^(-)])/[[HA]]`

From our ice table and the values given, we can plug all of the equilibrium values into the `K_a` expression as `K_a` is constant.

`(6.3 times10^-5)=(x^2)/(0.64-x)`

However, the change in concentration of the acid can be considered negligible, due to `K_a` being small: `(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)`

Hence:

`x=0.0063498031`

There the equation becomes:

`[H_3O^(+)]=x=0.0063498031`

`pH=-log[H_3O^(+)]`

`pH=-log[0.0063498031]`

`pH approx 2.2`