Chapter 16, Problem 16.21QP

(a)

Interpretation Introduction

Interpretation: From the given concentration of ${\text{OH}}^{\text{-}}$ ion in aqueous solution at $\text{100 }°\text{C}$, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has to be calculated.

Concept Introduction:

Autoionization of water is the reaction in which the water undergoes ionization to give a proton and a hydroxide ion.  Water is a very weak electrolyte and hence it does not completely dissociate into the ions.  The ionization happens to a very less extent only.  The ionization of water is an equilibrium reaction and hence this has equilibrium rate constant.

$K_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=1.0\times {10}^{-14}$

To calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Answer to Problem 16.21QP

Answer

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in (a) is $2.05\times {10}^{-11}M$

Explanation of Solution

Given

Concentration of ${\text{OH}}^{\text{-}}$ = $2.50\times {10}^{-2}M$

Formula

$\begin{array}{l}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=5.13\times {10}^{-13}\text{at 100 }°\text{C}\\ [{\text{H}}_{\text{3}}{\text{O}}^{+}]=\frac{5.13\times {10}^{-13}}{[{\text{OH}}^{-}]}\end{array}$

Where,

${\text{[OH}}^{\text{-}}\text{]}$ is concentration of ${\text{OH}}^{\text{-}}$

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Substitute the given concentration of ${\text{OH}}^{\text{-}}$ in the above formula,

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{5.13\times {10}^{-13}}{2.50\times {10}^{-2}}\\ =2.05\times {10}^{-11}M\end{array}$

Thus the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $2.05\times {10}^{-11}M$

(b)

Interpretation Introduction

Interpretation: From the given concentration of ${\text{OH}}^{\text{-}}$ ion in aqueous solution at $\text{100 }°\text{C}$, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has to be calculated.

Concept Introduction:

Autoionization of water is the reaction in which the water undergoes ionization to give a proton and a hydroxide ion.  Water is a very weak electrolyte and hence it does not completely dissociate into the ions.  The ionization happens to a very less extent only.  The ionization of water is an equilibrium reaction and hence this has equilibrium rate constant.

$K_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=1.0\times {10}^{-14}$

To calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Answer to Problem 16.21QP

Answer

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in (b) is $3.07\times {10}^{-8}M$

Explanation of Solution

Given

Concentration of ${\text{OH}}^{\text{-}}$ = $1.67\times {10}^{-5}M$

Formula

$\begin{array}{l}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=5.13\times {10}^{-13}\text{at 100 }°\text{C}\\ [{\text{H}}_{\text{3}}{\text{O}}^{+}]=\frac{5.13\times {10}^{-13}}{[{\text{OH}}^{-}]}\end{array}$

Where,

${\text{[OH}}^{\text{-}}\text{]}$ is concentration of ${\text{OH}}^{\text{-}}$

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Substitute the given concentration of ${\text{OH}}^{\text{-}}$ in the above formula,

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{5.13\times {10}^{-13}}{1.67\times {10}^{-5}}\\ =3.07\times {10}^{-8}M\end{array}$

Thus the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $3.07\times {10}^{-8}M$

c)

Interpretation Introduction

Interpretation: From the given concentration of ${\text{OH}}^{\text{-}}$ ion in aqueous solution at $\text{100 }°\text{C}$, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has to be calculated.

Concept Introduction:

Autoionization of water is the reaction in which the water undergoes ionization to give a proton and a hydroxide ion.  Water is a very weak electrolyte and hence it does not completely dissociate into the ions.  The ionization happens to a very less extent only.  The ionization of water is an equilibrium reaction and hence this has equilibrium rate constant.

$K_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=1.0\times {10}^{-14}$

To calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Answer to Problem 16.21QP

Answer

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in (c) is $5.95\times {10}^{-11}M$

Explanation of Solution

Given

Concentration of ${\text{OH}}^{\text{-}}$ = $8.62\times {10}^{-3}M$

Formula

$\begin{array}{l}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=5.13\times {10}^{-13}\text{at 100 }°\text{C}\\ [{\text{H}}_{\text{3}}{\text{O}}^{+}]=\frac{5.13\times {10}^{-13}}{[{\text{OH}}^{-}]}\end{array}$

Where,

${\text{[OH}}^{\text{-}}\text{]}$ is concentration of ${\text{OH}}^{\text{-}}$

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Substitute the given concentration of ${\text{OH}}^{\text{-}}$ in the above formula,

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{5.13\times {10}^{-13}}{8.62\times {10}^{-3}}\\ =5.95\times {10}^{-11}M\end{array}$

Thus the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $5.95\times {10}^{-11}M$

(d)

Interpretation Introduction

Interpretation: From the given concentration of ${\text{OH}}^{\text{-}}$ ion in aqueous solution at $\text{100 }°\text{C}$, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has to be calculated.

Concept Introduction:

Autoionization of water is the reaction in which the water undergoes ionization to give a proton and a hydroxide ion.  Water is a very weak electrolyte and hence it does not completely dissociate into the ions.  The ionization happens to a very less extent only.  The ionization of water is an equilibrium reaction and hence this has equilibrium rate constant.

$K_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=1.0\times {10}^{-14}$

To calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Answer to Problem 16.21QP

Answer

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in (d) is $2.93\times {10}^{-1}M$

Explanation of Solution

Given

Concentration of ${\text{OH}}^{\text{-}}$ = $1.75\times {10}^{-12}M$

Formula

$\begin{array}{l}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=5.13\times {10}^{-13}\text{at 100 }°\text{C}\\ [{\text{H}}_{\text{3}}{\text{O}}^{+}]=\frac{5.13\times {10}^{-13}}{[{\text{OH}}^{-}]}\end{array}$

Where,

${\text{[OH}}^{\text{-}}\text{]}$ is concentration of ${\text{OH}}^{\text{-}}$

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Substitute the given concentration of ${\text{OH}}^{\text{-}}$ in the above formula,

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{5.13\times {10}^{-13}}{1.75\times {10}^{-12}}\\ =2.93\times {10}^{-1}M\end{array}$

Thus the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $2.93\times {10}^{-1}M$