# A 0.100 M solution of an acid, HA, has a pH = 2.00. what is the value of the ionization constant, K_a for this acid?

Jan 1, 2017

${K}_{\text{a}} = 0.11$

#### Explanation:

Let's set up an ICE table to solve this problem.

$\textcolor{w h i t e}{m m m m m m m} \text{HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l} 0.100 \textcolor{w h i t e}{m m m m m l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mm)"-} x \textcolor{w h i t e}{m m m m m} + x \textcolor{w h i t e}{m l} + x$
$\text{E/mol·L"^"-1":color(white)(m)"0.100-} x \textcolor{w h i t e}{m m m m l l} x \textcolor{w h i t e}{m m l l} x$

We must use the $\text{pH}$ to calculate the value of $x$.

$\text{pH = 2.00}$

["H"_3"O"^"+"] = 10^"-2.00" color(white)(l)"mol/L" = "0.0100 mol/L" = x

The ${K}_{\text{a}}$ expression is:

K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x)

${K}_{\text{a}} = {x}^{2} / \left(0.100 - x\right) = {0.100}^{2} / \left(0.100 - 0.010\right) = \frac{0.0100}{0.090} = 0.11$