Chapter 18.3, Problem 42SSC

Interpretation Introduction

(a)

Interpretation:

$pH$ of the solution is to be found out using given concentration.

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:

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

Answer to Problem 42SSC

$pH=0$

Explanation of Solution

Given:

$\left[HI\right]=1M$

Therefore, $\left[H^{+}\right]=1M$

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

Therefore,

$\begin{array}{c}pH=-\log 1\\ =0\end{array}$

Interpretation Introduction

(b)

Interpretation:

$pH$ of the solution is to be found out using given concentration.

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:

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

Answer to Problem 42SSC

$pH=1.30$

Explanation of Solution

Given:

$\left[HN{O}_3\right]=0.05M$

Therefore, $\left[H^{+}\right]=0.05M$

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

Therefore,

$\begin{array}{c}pH=-\log 0.05\\ =-\log (5\times {10}^{-2})\\ =-\log 5-\log {10}^{-2}\\ =-0.70+2\\ =1.30\end{array}$

Interpretation Introduction

(c)

Interpretation:

$pH$ of the solution is to be found out using given concentration.

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:

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

Answer to Problem 42SSC

$pH=14$

Explanation of Solution

Given:

$\begin{array}{l}\left[KOH\right]=1M\\ \therefore \left[O{H}^{-}\right]=1M\end{array}$

$pOH=-\log \left[O{H}^{-}\right]$

Therefore,

$\begin{array}{c}pOH=-\log (1)\\ =0\end{array}$

Therefore,

$\begin{array}{c}pH=14-0\\ =14\end{array}$

Interpretation Introduction

(d)

Interpretation:

$pH$ of the solution is to be found out using given concentration.

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:

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

Answer to Problem 42SSC

$pH=9.68$

Explanation of Solution

Given:

$\begin{array}{l}\left[Mg{\left(OH\right)}_2\right]=2.4\times {10}^{-5}M\\ \therefore \left[O{H}^{-}\right]=2\times 2.4\times {10}^{-5}M\\ =4.8\times {10}^{-5}M\end{array}$

$pOH=-\log \left[O{H}^{-}\right]$

Therefore,

$\begin{array}{c}pOH=-\log (4.8\times {10}^{-5})\\ =-\log 4.9-\log {10}^{-5}\\ =-0.68+5\\ =4.32\end{array}$

Therefore,

$\begin{array}{c}pH=14-4.32\\ =9.68\end{array}$