Chapter 20, Problem 20.1P

Interpretation Introduction

Interpretation:

The concentrations of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{, }\left[{\text{ClO}}_{\text{4}}^{\text{-}}\right]$ and $\left[{\text{OH}}^{\text{-}}\right]$ in aqueous solution have to be calculated.

Concept Introduction:

$\text{Bronsted-Lowry Acids}$: If a species loses a proton then it is considered as $\text{Bronsted-Lowry acid}\text{.}$

Answer to Problem 20.1P

The concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ is $0.150\text{M}$.

The concentration of $\left[{\text{ClO}}_{\text{4}}^{\text{-}}\right]$ is $0.150\text{M}$.

The concentration of $\left[{\text{OH}}^{\text{-}}\right]$ is $6.67\times {10}^{-14}M$.

The given solution is acidic nature.

Explanation of Solution

Given: $0.150{\text{ M HClO}}_{\text{4}}$.

Generally, ${\text{HClO}}_{\text{4}}$ is strong acid since it completely dissociates into ions with easy donation of protons. Strong acid forms weaker conjugated base.

    ${\text{HClO}}_{\text{4}}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right){\text{+ClO}}_{\text{4}}^{\text{-}}\left(aq\right)$

1mole of ${\text{HClO}}_{\text{4}}\left(aq\right)$ gives 1 mole of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)$ and ${\text{ClO}}_{\text{4}}^{\text{-}}\left(aq\right)$. Hence, the concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=\left[{\text{ClO}}_{\text{4}}^{\text{-}}\right]\text{ = 0}\text{.150M}$.

Now, the hydroxide ion concentration is calculated.

The hydroxide ion concentration is determined by considering the ion product of water equation as follows,

  $\begin{array}{c}{K}_W=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\\ \\ 1.00\times {10}^{-14}M=0.150M\times \left[O{H}^{-}\right]\\ \\ \left[O{H}^{-}\right]=\frac{1.00\times {10}^{-14}M}{0.150M}\\ \\ =6.67\times {10}^{-14}M\end{array}$

The $\text{pH}$ of given aqueous solution is calculated as shown below,

    $\begin{array}{c}\text{pH = -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ =\text{ -log}\left[\text{0}\text{.150M}\right]\\ \\ \text{pH}=\text{ 0}\text{.824}\end{array}$

The above calculated $\text{pH}$ value is less than 7 which shows that the given aqueous solution is acidic nature.