Chapter 8, Problem 1MCP

(a)

Interpretation Introduction

Interpretation:

The number of moles of potassium hydroxide present in one litre solution.

Explanation of Solution

Given,

    $\text{3}\text{.00 g of KOH}$

The number of moles of potassium hydroxide is calculated as,

    $\begin{array}{l}\text{Moles}\text{of}\text{KOH}\text{=}\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\text{=}\frac{\text{3}\text{.00g}}{\text{56}\text{.1056 g/mol}}\\ \\ \text{=}\text{0}\text{.053mol}\text{KOH}\text{.}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The molarity of potassium hydroxide solution has to be calculated.

Concept Introduction:

Refer to part a.

Explanation of Solution

Given,

    $\text{3}\text{.00 g of KOH}$

The number of moles of potassium hydroxide is calculated as,

    $\begin{array}{l}\text{Moles}\text{of}\text{KOH}\text{=}\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\text{=}\frac{\text{3}\text{.00g}}{\text{56}\text{.1056 g/mol}}\\ \\ \text{=}\text{0}\text{.053mol}\text{KOH}\end{array}$

The molarity of potassium hydroxide solution is calculated as,

    $\begin{array}{l}\text{Molarity}\text{of}\text{KOH}\text{=}\frac{\text{0}\text{.053mol}\text{KOH}}{\text{1L}}\\ \\ \text{=}\text{0}\text{.053M}\text{.}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The molarity of hydroxide ion in solution has to be calculated.

Concept Introduction:

Refer to part a.

Explanation of Solution

Given,

    $\text{3}\text{.00 g of KOH}$

The number of moles of potassium hydroxide is calculated as,

    $\begin{array}{l}\text{Moles}\text{of}\text{KOH}\text{=}\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\text{=}\frac{\text{3}\text{.00g}}{\text{56}\text{.1056 g/mol}}\\ \\ \text{=}\text{0}\text{.053mol}\text{KOH}\end{array}$

The molarity of potassium hydroxide solution is calculated as,

    $\begin{array}{l}\text{Molarity}\text{of}\text{KOH}\text{=}\frac{\text{0}\text{.053mol}\text{KOH}}{\text{1L}}\\ \\ \text{=}\text{0}\text{.053M}\text{.}\end{array}$

Potassium hydroxide is a strong base which dissociates completely in aqueous solution.  The concentration of hydroxide ion is same as concentration of potassium hydroxide.  Therefore, molarity of hydroxide ion is $\text{0}\text{.053M}$

(d)

Interpretation Introduction

Interpretation:

The potassium hydroxide solution has to be classified as acidic, basic, or neutral solution.

Concept Introduction:

Acidic solution is a solution which contains more hydrogen ion in it.  Basic solution has more hydroxide ion concentration.  Neutral solution has equal amount of  hydroxide ion and hydronium ion in it.

Explanation of Solution

Potassium hydroxide dissociates completely in aqueous solution.  The dissociation of potassium hydroxide is given by

    ${\text{KOH}}_{\text{(aq)}}\stackrel{}{\to }{\text{K}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{OH}}^{\text{-}}{}_{\text{(aq)}}$

Potassium hydroxide gives hydroxide ion in aqueous solution.  Therefore, it is basic solution.

(e)

Interpretation Introduction

Interpretation:

The molarity of hydronium ion in solution has to be calculated.

Concept Introduction:

Refer to part a.

Explanation of Solution

Given,

    $\text{3}\text{.00 g of KOH}$

The molarity of potassium hydroxide solution is calculated as,

    $\begin{array}{l}\text{Moles}\text{of}\text{KOH}\text{=}\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\text{=}\frac{\text{3}\text{.00g}}{\text{56}\text{.1056 g/mol}}\\ \\ \text{=}\text{0}\text{.053mol}\text{KOH}\\ \\ \text{Molarity}\text{of}\text{KOH}\text{=}\frac{\text{0}\text{.053mol}\text{KOH}}{\text{1L}}\\ \\ \text{=}\text{0}\text{.053M}\text{.}\end{array}$

The molarity of hydronium ion is calculated as,

    $\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}\text{0}\text{.053M}\\ \\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{.00×}{\text{10}}^{\text{-14}}\\ \\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\frac{\text{1}\text{.00×}{\text{10}}^{\text{-14}}}{\text{0}\text{.053}}\text{=}\text{1}\text{.89}\text{×}{\text{10}}^{\text{-13}}\text{M}\text{.}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of potassium hydroxide solution has to be calculated.

Concept Introduction:

pH:  The $\text{pH}$ of a solution is defined as the negative logarithm of the molar concentration of the hydronium ion.

    $\text{pH}\text{=}\text{-log}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Explanation of Solution

Given,

    $\text{3}\text{.00 g of KOH}$

The molarity of potassium hydroxide solution is calculated as,

    $\begin{array}{l}\text{Moles}\text{of}\text{KOH}\text{=}\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\text{=}\frac{\text{3}\text{.00g}}{\text{56}\text{.1056 g/mol}}\\ \\ \text{=}\text{0}\text{.053mol}\text{KOH}\\ \\ \text{Molarity}\text{of}\text{KOH}\text{=}\frac{\text{0}\text{.053mol}\text{KOH}}{\text{1L}}\\ \\ \text{=}\text{0}\text{.053M}\end{array}$

The molarity of hydronium ion is calculated as,

    $\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}\text{0}\text{.053M}\\ \\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{.00×}{\text{10}}^{\text{-14}}\\ \\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\frac{\text{1}\text{.00×}{\text{10}}^{\text{-14}}}{\text{0}\text{.053}}\text{=}\text{1}\text{.89}\text{×}{\text{10}}^{\text{-13}}\text{M}\end{array}$

The $\text{pH}$ of potassium hydroxide solution is calculated as,

    $\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\text{1}\text{.89}\text{×}{\text{10}}^{\text{-13}}\text{M}\\ \\ \text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\text{-log(}\text{1}\text{.89}\text{×}{\text{10}}^{\text{-13}}\text{)}\\ \\ \text{=}\text{12}\text{.72}\end{array}$

(g)

Interpretation Introduction

Interpretation:

A compound that can be added with potassium hydroxide to form a neutral solution has to be identified.

Concept Introduction:

Refer to part d.

Explanation of Solution

A neutral solution is formed when a strong base reacts with strong acid.  Hydrochloric acid is a strong acid that can react with potassium hydroxide to give a neutral solution.

    ${\text{HCl}}_{\text{(aq)}}\text{+}{\text{KOH}}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{+}{\text{KCl}}_{\text{(aq)}}$

(h)

Interpretation Introduction

Interpretation:

The balanced, net, and ionic chemical equations between potassium hydroxide and hydrochloric acid have to be given.

Explanation of Solution

A neutral solution is formed when a strong base reacts with strong acid.  Hydrochloric acid is a strong acid that can react with potassium hydroxide to give a neutral solution.

    ${\text{HCl}}_{\text{(aq)}}\text{+}{\text{KOH}}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{+}{\text{KCl}}_{\text{(aq)}}$

The complete ionic equation is given by

    ${\text{H}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{Cl}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{K}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{K}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{Cl}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The net ionic equation is given by

    ${\text{H}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$