Chapter 16, Problem 16.20QP

(a)

Interpretation Introduction

Interpretation: From the given concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion in aqueous solution at $\text{50 }°\text{C}$, the concentration of ${\text{OH}}^{\text{-}}$ 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 ${\text{OH}}^{\text{-}}$

Answer to Problem 16.20QP

Answer

The concentration of ${\text{OH}}^{\text{-}}$ in (a) is $4.85\times {10}^{-10}M$

Explanation of Solution

Given

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ = $1.13\times {10}^{-4}M$

Formula

$\begin{array}{l}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=5.84\times {10}^{-14}\text{at 50 }°\text{C}\\ [{\text{OH}}^{-}]=\frac{5.84\times {10}^{-14}}{[{\text{H}}_{\text{3}}{\text{O}}^{+}]}\end{array}$

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 ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in the above formula,

$\begin{array}{l}[{\text{OH}}^{-}]=\frac{5.84\times {10}^{-14}}{1.13\times {10}^{-4}}\\ =4.85\times {10}^{-10}M\end{array}$

Thus the concentration of ${\text{OH}}^{\text{-}}$ is $4.85\times {10}^{-10}M$

(b)

Interpretation Introduction

Interpretation: From the given concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion in aqueous solution at $\text{50 }°\text{C}$, the concentration of ${\text{OH}}^{\text{-}}$ 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 ${\text{OH}}^{\text{-}}$

Answer to Problem 16.20QP

Answer

The concentration of ${\text{OH}}^{\text{-}}$ in (b) is $1.2\times {10}^{-6}M$

Explanation of Solution

Given

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ = $4.55\times {10}^{-8}M$

Formula

$\begin{array}{l}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=5.84\times {10}^{-14}\text{at 50 }°\text{C}\\ [{\text{OH}}^{-}]=\frac{5.84\times {10}^{-14}}{[{\text{H}}_{\text{3}}{\text{O}}^{+}]}\end{array}$

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 ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in the above formula,

$\begin{array}{l}[{\text{OH}}^{-}]=\frac{5.84\times {10}^{-14}}{4.55\times {10}^{-8}}\\ =1.20\times {10}^{-6}M\end{array}$

Thus the concentration of ${\text{OH}}^{\text{-}}$ is $1.20\times {10}^{-6}M$

(c)

Interpretation Introduction

Interpretation: From the given concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion in aqueous solution at $\text{50 }°\text{C}$, the concentration of ${\text{OH}}^{\text{-}}$ 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 ${\text{OH}}^{\text{-}}$

Answer to Problem 16.20QP

Answer

The concentration of ${\text{OH}}^{\text{-}}$ in (c) is $7.77\times {10}^{-4}M$

Explanation of Solution

Given

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ = $7.05\times {10}^{-11}M$

Formula

$\begin{array}{l}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=5.84\times {10}^{-14}\text{at 50 }°\text{C}\\ [{\text{OH}}^{-}]=\frac{5.84\times {10}^{-14}}{[{\text{H}}_{\text{3}}{\text{O}}^{+}]}\end{array}$

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 ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in the above formula,

$\begin{array}{l}[{\text{OH}}^{-}]=\frac{5.84\times {10}^{-14}}{7.05\times {10}^{-11}}\\ =7.77\times {10}^{-4}M\end{array}$

Thus the concentration of ${\text{OH}}^{\text{-}}$ is $7.77\times {10}^{-4}M$

(d)

Interpretation Introduction

Interpretation: From the given concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion in aqueous solution at $\text{50 }°\text{C}$, the concentration of ${\text{OH}}^{\text{-}}$ 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 ${\text{OH}}^{\text{-}}$

Answer to Problem 16.20QP

Answer

The concentration of ${\text{OH}}^{\text{-}}$ in (d) is $1.75\times {10}^{-12}M$

Explanation of Solution

Given

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ = $3.13\times {10}^{-2}M$

Formula

$\begin{array}{l}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=5.84\times {10}^{-14}\text{at 50 }°\text{C}\\ [{\text{OH}}^{-}]=\frac{5.84\times {10}^{-14}}{[{\text{H}}_{\text{3}}{\text{O}}^{+}]}\end{array}$

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 ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in the above formula,

$\begin{array}{l}[{\text{OH}}^{-}]=\frac{5.84\times {10}^{-14}}{3.13\times {10}^{-2}}\\ =1.75\times {10}^{-12}M\end{array}$

Thus the concentration of ${\text{OH}}^{\text{-}}$ is $1.75\times {10}^{-12}M$