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{0.115M}}$$\mathbf{\text{HClO}}$ ?

Short 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$\text{[HClO]=0.115M}$

Step-by-step solution

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.

$\begin{array}{l}\left[\text{HClO}\right]{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{+}\left[{\text{ClO}}^{\text{-}}\right]\\ \text{I0.115M00}\\ \text{C-x+x+x}\\ \text{E0.115M-x+x+x}\\ \\ {\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{ClO}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[\text{HClO}\right]}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{1.25-x}}\end{array}$

$\text{x=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{ClO}}^{\text{-}}\right]$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 $\text{0.115M}$. 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.

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

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$\text{[HClO]=0.115M}$