Chapter 16, Problem 42A

(a)

Interpretation Introduction

Interpretation: The hydrogen ion concentration and $pH$ of the HCl 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 42A

  $pH=4$ and $\left[H^{+}\right]=0.00010M$ .

Explanation of Solution

  $\left[HCl\right]=0.00010M$

Strong acids are completely ionized.

Therefore,

  $\left[H^{+}\right]=0.00010M$

  $pH=-\log \left[H^{+}\right]$

Therefore,

  $\begin{array}{c}pH=-\log 0.00010\\ =-\log (1\times {10}^{-4})\\ =4\end{array}$

(b)

Interpretation Introduction

Interpretation:

The hydrogen ion concentration and $pH$ of the HNO₃ 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 42A

  $pH=2.301$ and $\left[H^{+}\right]=0.0050M$ .

Explanation of Solution

  $\left[HN{O}_3\right]=0.0050M$

Strong acids are completely ionized.

Therefore,

  $\left[H^{+}\right]=0.0050M$

  $pH=-\log \left[H^{+}\right]$

Therefore,

  $\begin{array}{c}pH=-\log 0.0050\\ =-\log (5\times {10}^{-3})\\ =-\log 5-\log {10}^{-3}\\ =-0.699+3\\ =2.301\end{array}$

(c)

Interpretation Introduction

Interpretation:

The hydrogen ion concentration and $pH$ of the HClO₃ 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 42A

  $pH=4.376$

  $\left[H^{+}\right]=4.21\times {10}^{-5}M$

Explanation of Solution

  $\left[HCl{O}_4\right]=4.21\times {10}^{-5}M$

Strong acids are completely ionized.

Therefore,

  $\left[H^{+}\right]=4.21\times {10}^{-5}M$

  $pH=-\log \left[H^{+}\right]$

Therefore,

  $\begin{array}{c}pH=-\log (4.21\times {10}^{-5})\\ =-\log 4.21-\log {10}^{-5}\\ =-0.624+5\\ =4.376\end{array}$

(d)

Interpretation Introduction

Interpretation:

The hydrogen ion concentration and $pH$ of the HNO₃ 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 42A

  $pH=2.199$

  $\left[H^{+}\right]=6.33\times {10}^{-3}M$ .

Explanation of Solution

Given:

  $\left[HN{O}_3\right]=6.33\times {10}^{-3}M$

Strong acids are completely ionized.

Therefore,

  $\left[H^{+}\right]=6.33\times {10}^{-3}M$

  $pH=-\log \left[H^{+}\right]$

Therefore,

  $\begin{array}{c}pH=-\log (6.33\times {10}^{-3})\\ =-\log 6.33-\log {10}^{-3}\\ =-0.801+3\\ =2.199\end{array}$