# Question #a3dd1

May 29, 2015

Your solution will have a pH equal to 4.0.

Since you're dealing with a weak acid, you can use an ICE table on its dissociation equilibrium to determine the equilibrium concentrations of the species involved in the reaction

$\text{ } H {A}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + {A}_{\left(a q\right)}^{-}$
I.......1.........................................0.................0
C....(-x)......................................(+x).............(+x)
E.....1-x........................................x.................x

The acid dissociation constant, ${K}_{a}$, will be equal to

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[{A}^{-}\right]}{\left[H A\right]} = \frac{x \cdot x}{1 - x} = {x}^{2} / \left(1 - x\right)$

Since ${K}_{a}$ has such a small value, you ca approximate (1-x) to be equal to 1, which would mean that

${K}_{a} = {x}^{2} / 1 = {x}^{2} = {10}^{- 8} \implies x = \sqrt{{10}^{- 8}} = {10}^{- 4}$

Since$x$ is equal to the equilibrium concentration of the hydronium ion, you'll have

$\left[{H}_{3} {O}^{+}\right] = {10}^{- 4} \text{M}$

Therefore, the pH of the solution will be

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

$p {H}_{\text{sol}} = - \log \left({10}^{- 4}\right) = \textcolor{g r e e n}{4.0}$