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) ?

Apr 16, 2018

See below:

Explanation:

Start by setting up an ICE table:

We have the following reaction:

$H A \left(a q\right) + {H}_{2} O \left(a q\right) r i g h t \le f t h a r p \infty n s {A}^{-} \left(a q\right) + {H}_{3} {O}^{+} \left(a q\right)$

And we have an initial concentration of $H A$ at 0.64 $m o l {\mathrm{dm}}^{-} 3$, so let's plug what we have into the ICE table:

$\textcolor{w h i t e}{m m m m m i} H A \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {A}^{-} \left(a q\right) + {H}_{3} {O}^{+} \left(a q\right)$

$\text{Initial:} \textcolor{w h i t e}{m m} 0.64 \textcolor{w h i t e}{m i i m m} - \textcolor{w h i t e}{m m m m m} 0 \textcolor{w h i t e}{m m m m m m} 0$

$\text{Change:} \textcolor{w h i t e}{i m} - x \textcolor{w h i t e}{m i i m m} - \textcolor{w h i t e}{m m m m} + x \textcolor{w h i t e}{m m m m i i} + x$

$\text{Eq:} \textcolor{w h i t e}{m m m} 0.64 - x \textcolor{w h i t e}{i i m m} - \textcolor{w h i t e}{m m m m m} x \textcolor{w h i t e}{m m m m m m} x$

Now using the ${K}_{a}$ expression:

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[{A}^{-}\right]}{\left[H A\right]}$

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.

$\left(6.3 \times {10}^{-} 5\right) = \frac{{x}^{2}}{0.64 - x}$

However, the change in concentration of the acid can be considered negligible, due to ${K}_{a}$ being small: $\left(0.64 - x = 0.64\right)$

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.

$\left(6.3 \times {10}^{-} 5\right) = \frac{{x}^{2}}{0.64}$

Hence:

$x = 0.0063498031$

There the equation becomes:

$\left[{H}_{3} {O}^{+}\right] = x = 0.0063498031$

$p H = - \log \left[{H}_{3} {O}^{+}\right]$

$p H = - \log \left[0.0063498031\right]$

$p H \approx 2.2$