Question

What would be the hydrogen ion concentration in a solution containing 0.04 mole of acetic acid and 0.05 mole of sodium acetate in 500 ml of the solution?
Dissociation constant for acetic acid is 1.8 x ${10}^{-5}$.
The [ $H^{+}$] is:

• A. 1.44 $\times$ ${10}^{-5}$
• B. 14.5 $\times$ ${10}^{-9}$
• C. 3.88 $\times$ ${10}^{-2}$
• D. none of the above

Answer 1

The correct option is A

1.44 $\times$ ${10}^{-5}$

Let us recall common ion effect,

The concentration of $C{H}_{3}COOH$ in the solution is = 0.04 $\times$ 2 = 0.08 M
The concentration of $C{H}_{3}COONa$ in the solution is = 0.05 $\times$ 2 = 0.10M
$C{H}_{3}COOH$ $\leftrightharpoons$ ${C{H}_{3}COO}^{-}$ + $H^{+}$

0.08 M $xM$

$C{H}_{3}COONa$ $\to$ ${C{H}_{3}COO}^{-}$ + ${Na}^{+}$

0.10M $0.10M$

$K_a$ = $\frac{\left[H^{+}\right]\left[{C{H}_{3}COO}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

From the first reaction there will be some ${C{H}_{3}COO}^{-}$ ions. But look at the 0.1 M which we get from the weak acid + strong base salt. That would be much larger than the [${C{H}_{3}COO}^{-}$] we get from the weak acid alone. So as an approximation, we discard acid's acetate ion contribution and just take [${C{H}_{3}COO}^{-}$] from salt = 0.1 M

$Ka=\frac{\left[H^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

$\left[H^{+}\right]=\frac{Ka\left[C{H}_{3}COOH\right]}{\left[C{H}_{3}CO{O}^{-}\right]}$
=$\frac{1.8\times {10}^{-5}\times 0.08}{0.10}$
=$1.44\times {10}^{-5}$

$K_a$[${C{H}_{3}COO}^{-}$] and $\left[C{H}_{3}COOH\right]$ are all known. Solving for $x$ or for [ $H^{+}$] , we get 1.44 x ${10}^{-5}$