# Question 9ff52

Jun 9, 2017

$\text{pH} = 3.32$

#### Explanation:

The idea here is that a weak acid is only partially ionized in aqueous solution, which means that in order to find the solution's concentration of hydronium ions, you must use the acid dissociation constant, ${K}_{a}$.

In this case, you have

${\text{HX"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "X}}_{\left(a q\right)}^{-}$

By definition, the acid dissociation constant is equal to

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["X"^(-)])/(["HX}\right]\right)$

Now, you can calculate the equilibrium concentrations of the three chemical species by using an ICE table

${\text{ ""HX"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) " "+" " "X}}_{\left(a q\right)}^{-}$

$\textcolor{p u r p \le}{\text{I}} \textcolor{w h i t e}{a a a a \textcolor{b l a c k}{0.23} a a a a a a a a a a a a a a a a \textcolor{b l a c k}{0} a a a a a a a a a a a \textcolor{b l a c k}{0}}$
$\textcolor{p u r p \le}{\text{C}} \textcolor{w h i t e}{a a a \textcolor{b l a c k}{\left(- x\right)} a a a a a a a a a a a a a \textcolor{b l a c k}{\left(+ x\right)} a a a a a a a \textcolor{b l a c k}{\left(+ x\right)}}$
$\textcolor{p u r p \le}{\text{E}} \textcolor{w h i t e}{a a \textcolor{b l a c k}{0.23 - x} a a a a a a a a a a a a a a \textcolor{b l a c k}{x} a a a a a a a a a a a \textcolor{b l a c k}{x}}$

The expression of the acid dissociation constant can thus be written as

${K}_{a} = \frac{x \cdot x}{0.23 - x}$

which is equivalent to

$1.0 \cdot {10}^{- 6} = {x}^{2} / \left(0.23 - x\right)$

Now, because the acid dissociation constant is very small compared to the initial concentration of the acid, you can use the approximation

$0.23 - x \approx 0.23$

This means that you have

$1.0 \cdot {10}^{- 6} = {x}^{2} / 0.23$

Solve for $x$ to get

$x = \sqrt{0.23 \cdot 1.0 \cdot {10}^{- 6}} = 0.0004796$

Since $x$ represents the equilibrium concentration of hydronium cations, you will have

["H"_3"O"^(+)] = "0.0004796 M"

Consequently, the $\text{pH}$ of the solution

"pH" = - log(["H"_3"O"^(+)])#

will be equal to

$\text{pH} = - \log \left(0.0004796\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{3.32}}}$

The answer is rounded to two decimal places, the number of sig figs you have for the initial concentration of the acid.