# What is the original molarity of a solution of a weak acid, given Ka and pH?

## ${K}_{a} = 3.5 \cdot {10}^{-} 5$ $p H = 5.25$ @25 degrees Celsius

Mar 8, 2018

$6.5 \cdot {10}^{-} 6$ M

#### Explanation:

Construct an ICE table using the following reaction equation:
${H}_{2} O$ + $H A$ $r i g h t \le f t h a r p \infty n s$ ${A}^{-}$ + ${H}_{3} {O}^{+}$

Use the pH to calculate $\left[{H}_{3} {O}^{+}\right]$ at equilibrium, which is also the change in concentration for the table.

Equilibrium Concentrations:
$\left[H A\right] = x - 5.6 \cdot {10}^{-} 6$ M
$\left[{A}^{-}\right] = 5.6 \cdot {10}^{-} 6$ M
$\left[{H}_{3} {O}^{+}\right] = 5.6 \cdot {10}^{-} 6$ M

Set up an equilibrium expression using ${K}_{a}$:
$3.5 \cdot {10}^{-} 5 = {\left(5.6 \cdot {10}^{-} 6\right)}^{2} / \left(x - 5.6 \cdot {10}^{-} 6\right)$
$x = 9.0 \cdot {10}^{-} 7$

$\left[H A\right] = 9.0 \cdot {10}^{-} 7$ M $- 5.6 \cdot {10}^{-} 6$ M $= 6.5 \cdot {10}^{-} 6$ M