Q18.68P - Chemistry: Molecular Nature Of Matter And Change

Question

Hypochlorous acid, $\mathbf{\text{HClO}}$, has a ${\mathbf{\text{pK}}}_{\mathbf{\text{a}}}$ of $\mathbf{\text{7.54}}$. What are $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$, $\mathbf{\text{pH}}$, $\left[{\text{ClO}}^{\text{-}}\right]$, and $\left[\text{HClO}\right]$ in $\mathbf{\text{0.115 M}}$ $\mathbf{\text{HClO}}$ ?

Step-by-Step Solution

Answer

Values are:

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }\left[{\text{ClO}}^{\text{-}}\right]\\ {\text{= 5.75 × 10}}^{\text{-5}}\text{ M}\end{array}$

$\text{pH = 4.24 }$and $\text{[HClO] = 0.115 M}$

Define acids

Chemical agents that release hydrogen ions when introduced to water are referred to as acids in chemistry.

Explanation

To begin, find ${\text{pK}}_{\text{a}}$

$\begin{array}{c}{\text{pK}}_{\text{a}}{\text{ = -log(K}}_{\text{a}}\text{)}\\ {\text{K}}_{\text{a}}{\text{ = 10}}^{{\text{-pK}}_{\text{a}}}\\ {\text{= 10}}^{\text{-7.54}}\\ {\text{K}}_{\text{a}}{\text{ = 2.88 × 10}}^{\text{-8}}\end{array}$

Then, for the dissociation of $\text{HCIO}$, construct the reaction equation.

${\text{HCIO + H}}_{\text{2}}\text{O }\rightleftharpoons {\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + ClO}}^{\text{-}}$

After that, create the ICE table to get the ${\text{K}}_{\text{a}}$ equation.

$\begin{array}{l}\left[\text{HClO}\right]{\text{ + H}}_{\text{2}}\text{O }\rightleftharpoons \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ + }\left[{\text{ClO}}^{\text{-}}\right]\\ \text{I 0.115 M 0 0}\\ \text{C -x +x +x}\\ \text{E 0.115M-x +x +x}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{ClO}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[\text{HClO}\right]}\\ {\text{K}}_{\text{a}}\text{ = }\frac{{\text{x}}^{\text{2}}}{\text{1.25-x}}\end{array}$

$\text{x = }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }\left[{\text{ClO}}^{\text{-}}\right]$ is known. Since $\text{HClO}$ is such a weak acid, it ${\text{K}}_{\text{a}}$ must be extremely low. Assume that $\text{x}$ has no effect on the denominator's $\text{0.115 M}$. Then, to solve for $\text{x}$, substitute the ${\text{K}}_{\text{a}}$.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{ = }\frac{{\text{x}}^{\text{2}}}{\text{0.115}}\\ {\text{x}}^{\text{2}}{\text{ = (K}}_{\text{a}}\text{)}\left(\text{0.115}\right)\\ \text{x = }\sqrt{{\text{(K}}_{\text{a}}\text{)}\left(\text{0.115}\right)}\\ \text{= }\sqrt{{\text{(2.88×10}}^{\text{-8}}\text{)(0.115)}}\\ {\text{x = 5.75 × 10}}^{\text{-5}}\end{array}$

As, $\text{x = }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }\left[{\text{ClO}}^{\text{-}}\right]$

Then 

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }\left[{\text{ClO}}^{\text{-}}\right]\\ {\text{= 5.75 × 10}}^{\text{-5}}\text{ M}\end{array}$

The $\text{pH}$ of the solution should then be calculated.

$\begin{array}{c}\text{pH = -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{= -log}\left({\text{5.75 × 10}}^{\text{-5}}\right)\\ \text{pH = 4.24 }\end{array}$

Finally, find the final concentration of ${\text{ClCH}}_{\text{3}}\text{COOH}$ by solving.

$\begin{array}{c}\text{[HClO] = 0.115 - x}\\ {\text{= 0.115 - 5.75 × 10}}^{\text{-5}}\\ \text{[HClO] = 0.115 M}\end{array}$

Therefore, 

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }\left[{\text{ClO}}^{\text{-}}\right]\\ {\text{= 5.75 × 10}}^{\text{-5}}\text{ M}\end{array}$

$\text{pH = 4.24 }$and $\text{[HClO] = 0.115 M}$