Chapter 18.3, Problem 30PP

Interpretation Introduction

(a)

Interpretation:

$\left[H^{+}\right]and\left[O{H}^{-}\right]$ of the solution is to be found out using given $pH$.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ \left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}\\ \left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 30PP

$\left[H^{+}\right]=3.2\times {10}^{-7}M$

$\left[O{H}^{-}\right]=3.12\times {10}^{-8}M$.

Explanation of Solution

Given:

$pH=6.50$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=6.50\\ -\log \left[H^{+}\right]=6.50\\ \left[H^{+}\right]=Anti\log (-6.50)\\ =3.2\times {10}^{-7}M\end{array}$

Therefore,

$\begin{array}{c}\left[O{H}^{-}\right]=\frac{{10}^{-14}}{3.2\times {10}^{-7}}\\ =3.12\times {10}^{-8}M\end{array}$

Interpretation Introduction

(b)

Interpretation:

$\left[H^{+}\right]and\left[O{H}^{-}\right]$ of the solution is to be found out using given $pH$.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ \left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}\\ \left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 30PP

$\left[H^{+}\right]=4.3\times {10}^{-3}M$

$\left[O{H}^{-}\right]=2.3\times {10}^{-12}M$.

Explanation of Solution

Given:

$pH=2.37$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=2.37\\ -\log \left[H^{+}\right]=2.37\\ \left[H^{+}\right]=Anti\log (-2.37)\\ =4.3\times {10}^{-3}M\end{array}$

Therefore,

$\begin{array}{c}\left[O{H}^{-}\right]=\frac{{10}^{-14}}{4.3\times {10}^{-3}}\\ =2.3\times {10}^{-12}M\end{array}$

Interpretation Introduction

(c)

Interpretation:

$\left[H^{+}\right]and\left[O{H}^{-}\right]$ of the solution is to be found out using given $pH$.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ \left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}\\ \left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 30PP

$\left[H^{+}\right]=3.2\times {10}^{-11}M$

$\left[O{H}^{-}\right]=3.12\times {10}^{-4}M$.

Explanation of Solution

Given:

$pH=10.50$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=10.50\\ -\log \left[H^{+}\right]=10.50\\ \left[H^{+}\right]=Anti\log (-10.50)\\ =3.2\times {10}^{-11}M\end{array}$

Therefore,

$\begin{array}{c}\left[O{H}^{-}\right]=\frac{{10}^{-14}}{3.2\times {10}^{-11}}\\ =3.12\times {10}^{-12}M\end{array}$

Interpretation Introduction

(d)

Interpretation:

$\left[H^{+}\right]and\left[O{H}^{-}\right]$ of the solution is to be found out using given $pH$.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ \left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}\\ \left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 30PP

$\left[H^{+}\right]=1.3\times {10}^{-12}M$

$\left[O{H}^{-}\right]=7.9\times {10}^{-3}M$.

Explanation of Solution

Given:

$pH=11.90$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=11.90\\ -\log \left[H^{+}\right]=11.90\\ \left[H^{+}\right]=Anti\log (-11.90)\\ =1.3\times {10}^{-12}M\end{array}$

Therefore,

$\begin{array}{c}\left[O{H}^{-}\right]=\frac{{10}^{-14}}{1.3\times {10}^{-12}}\\ =7.9\times {10}^{-3}M\end{array}$