Chapter 18.3, Problem 22PP

(a)

Interpretation Introduction

Interpretation:

For the given solution $\left[{\text{H}}^{+}\right]$ and $\left[\text{O}{\text{H}}^{-}\right]$ are to be calculated and inference is to be drawn about its acidic and basic nature.

Concept introduction:

The given problem is based on the ionic product of water.

In an aqueous solution at 298 K,

$\begin{array}{l}{\text{K}}_{\text{w}}=\left[{\text{H}}^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]\end{array}$

  $\begin{array}{c}\text{w}\text{h}\text{e}\text{r}\text{e}\end{array}$

  $\begin{array}{c}\left[{\text{H}}^{+}\right]\text{i}\text{s}\text{co}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{i}\text{o}\text{n}\text{o}\text{f}{\text{H}}^{+}\text{i}\text{o}\text{n}\text{s}\end{array}$

  $\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]\text{i}\text{s}\text{co}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{i}\text{o}\text{n}\text{o}\text{f}\text{O}{\text{H}}^{-}\text{i}\text{o}\text{n}\text{s}\end{array}$

  $\begin{array}{c}{\text{K}}_{\text{w}}\text{i}\text{s}\text{i}\text{o}\text{n}\text{i}\text{c}\text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}\text{o}\text{f}\text{w}\text{a}\text{t}\text{e}\text{r}\end{array}$

The value of ${\text{K}}_{\text{w}}$ at 298 K is $1\times {10}^{-14}$ .

Answer to Problem 22PP

$\left[\text{O}{\text{H}}^{-}\right]={10}^{-1}\text{M}$

Explanation of Solution

${\text{K}}_{\text{w}}=\left[{\text{H}}^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]$

Given data:

$\left[{\text{H}}^{+}\right]=1\times {10}^{-13}\text{M}$

Therefore,

$\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]=\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}^{+}\right]}\end{array}$

   $\begin{array}{c}=\frac{1\times {10}^{-14}}{1\times {10}^{-13}}\end{array}$

   $\begin{array}{c}={10}^{-1}\text{M}\end{array}$

As $\left[\text{O}{\text{H}}^{-}\right]>\left[{\text{H}}^{+}\right]$

Therefore, solution is basic.

(b)

Interpretation Introduction

Interpretation:

For the given solution $\left[{\text{H}}^{+}\right]$ and $\left[\text{O}{\text{H}}^{-}\right]$ are to be calculated and inference is to be drawn about its acidic and basic nature.

Concept introduction:

The given problem is based on the ionic product of water.

In an aqueous solution at 298 K,

$\begin{array}{l}{\text{K}}_{\text{w}}=\left[{\text{H}}^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]\end{array}$

  $\begin{array}{c}\text{w}\text{h}\text{e}\text{r}\text{e}\end{array}$

  $\begin{array}{c}\left[{\text{H}}^{+}\right]\text{i}\text{s}\text{co}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{i}\text{o}\text{n}\text{o}\text{f}{\text{H}}^{+}\text{i}\text{o}\text{n}\text{s}\end{array}$

  $\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]\text{i}\text{s}\text{co}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{i}\text{o}\text{n}\text{o}\text{f}\text{O}{\text{H}}^{-}\text{i}\text{o}\text{n}\text{s}\end{array}$

  $\begin{array}{c}{\text{K}}_{\text{w}}\text{i}\text{s}\text{i}\text{o}\text{n}\text{i}\text{c}\text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}\text{o}\text{f}\text{w}\text{a}\text{t}\text{e}\text{r}\end{array}$

The value of ${\text{K}}_{\text{w}}$ at 298 K is $1\times {10}^{-14}$ .

Answer to Problem 22PP

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

Explanation of Solution

${\text{K}}_{\text{w}}=\left[{\text{H}}^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]$

Given data:

$\left[\text{O}{\text{H}}^{-}\right]=1\times {10}^{-7}\text{M}$

Therefore,

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{{\text{K}}_{\text{w}}}{\left[\text{O}{\text{H}}^{-}\right]}\end{array}$

   $\begin{array}{c}=\frac{1\times {10}^{-14}}{1\times {10}^{-7}}\end{array}$

   $\begin{array}{c}={10}^{-7}\text{M}\end{array}$

As $\left[\text{O}{\text{H}}^{-}\right]=\left[{\text{H}}^{+}\right]$

Therefore, solution is neutral.

(c)

Interpretation Introduction

Interpretation:

For the given solution $\left[{\text{H}}^{+}\right]$ and $\left[\text{O}{\text{H}}^{-}\right]$ are to be calculated and inference is to be drawn about its acidic and basic nature.

Concept introduction:

The given problem is based on the ionic product of water.

In an aqueous solution at 298 K,

$\begin{array}{l}{\text{K}}_{\text{w}}=\left[{\text{H}}^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]\end{array}$

  $\begin{array}{c}\text{w}\text{h}\text{e}\text{r}\text{e}\end{array}$

  $\begin{array}{c}\left[{\text{H}}^{+}\right]\text{i}\text{s}\text{co}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{i}\text{o}\text{n}\text{o}\text{f}{\text{H}}^{+}\text{i}\text{o}\text{n}\text{s}\end{array}$

  $\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]\text{i}\text{s}\text{co}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{i}\text{o}\text{n}\text{o}\text{f}\text{O}{\text{H}}^{-}\text{i}\text{o}\text{n}\text{s}\end{array}$

  $\begin{array}{c}{\text{K}}_{\text{w}}\text{i}\text{s}\text{i}\text{o}\text{n}\text{i}\text{c}\text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}\text{o}\text{f}\text{w}\text{a}\text{t}\text{e}\text{r}\end{array}$

The value of ${\text{K}}_{\text{w}}$ at 298 K is $1\times {10}^{-14}$ .

Answer to Problem 22PP

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

Explanation of Solution

${\text{K}}_{\text{w}}=\left[{\text{H}}^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]$

Given data:

$\left[\text{O}{\text{H}}^{-}\right]=1\times {10}^{-3}\text{M}$

Therefore,

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{{\text{K}}_{\text{w}}}{\left[\text{O}{\text{H}}^{-}\right]}\end{array}$

   $\begin{array}{c}=\frac{1\times {10}^{-14}}{1\times {10}^{-3}}\end{array}$

   $\begin{array}{c}={10}^{-11}\text{M}\end{array}$

As $\left[\text{O}{\text{H}}^{-}\right]>\left[{\text{H}}^{+}\right]$

Therefore, solution is basic.

(d)

Interpretation Introduction

Interpretation:

For the given solution $\left[{\text{H}}^{+}\right]$ and $\left[\text{O}{\text{H}}^{-}\right]$ are to be calculated and inference is to be drawn about its acidic and basic nature.

Concept introduction:

The given problem is based on the ionic product of water.

In an aqueous solution at 298 K,

$\begin{array}{l}{\text{K}}_{\text{w}}=\left[{\text{H}}^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]\end{array}$

  $\begin{array}{c}\text{w}\text{h}\text{e}\text{r}\text{e}\end{array}$

  $\begin{array}{c}\left[{\text{H}}^{+}\right]\text{i}\text{s}\text{co}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{i}\text{o}\text{n}\text{o}\text{f}{\text{H}}^{+}\text{i}\text{o}\text{n}\text{s}\end{array}$

  $\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]\text{i}\text{s}\text{co}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{i}\text{o}\text{n}\text{o}\text{f}\text{O}{\text{H}}^{-}\text{i}\text{o}\text{n}\text{s}\end{array}$

  $\begin{array}{c}{\text{K}}_{\text{w}}\text{i}\text{s}\text{i}\text{o}\text{n}\text{i}\text{c}\text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}\text{o}\text{f}\text{w}\text{a}\text{t}\text{e}\text{r}\end{array}$

The value of ${\text{K}}_{\text{w}}$ at 298 K is $1\times {10}^{-14}$ .

Answer to Problem 22PP

$\left[\text{O}{\text{H}}^{-}\right]=2.5\times {10}^{-10}\text{M}$

Explanation of Solution

${\text{K}}_{\text{w}}=\left[{\text{H}}^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]$

Given data:

$\left[{\text{H}}^{+}\right]=4\times {10}^{-5}\text{M}$

Therefore,

$\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]=\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}^{+}\right]}\end{array}$

   $\begin{array}{c}=\frac{1\times {10}^{-14}}{4\times {10}^{-5}}\end{array}$

   $\begin{array}{c}=2.5\times {10}^{-10}\text{M}\end{array}$

As $\left[{\text{H}}^{+}\right]>\left[\text{O}{\text{H}}^{-}\right]$

Therefore, solution is acidic.