Chapter 17, Problem 20CR

Interpretation Introduction

(a)

Interpretation:

The $\text{pH}$ and $\text{pOH}$ values for $0.00562\text{M}{\text{HClO}}_4$ solution are to be calculated.

Concept Introduction:

The term $\text{pH}$ referred to the power of hydrogen. The $\text{pH}$ of a solution is mathematically equal to the negative logarithm of concentration of ${\text{H}}^{+}$ ion in solution. The $\text{pH}$ of a solution is represented as,

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$

Where,

• $\left[{\text{H}}^{+}\right]$ represents the concentration of ${\text{H}}^{+}$ ions in the solution.

The $\text{pOH}$ of a solution is used to measure the alkalinity of the solution. The $\text{pOH}$ of a solution is mathematically equal to the negative logarithm of concentration of ${\text{OH}}^{-}$ ion in solution. The $\text{pOH}$ of a solution is represented as,

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ represents the concentration of ${\text{OH}}^{-}$ ions in the solution.

Answer to Problem 20CR

The $\text{pH}$ and $\text{pOH}$ values for $0.00562\text{M}{\text{HClO}}_4$ solution are $2.2503$ and $11.7497$ respectively.

Explanation of Solution

The concentration of the ${\text{HClO}}_4$ solution is $0.00562\text{M}$.

The dissociation of ${\text{HClO}}_4$ in aqueous medium is represented as,

${\text{HClO}}_4\left(\text{a}\text{q}\right)\to {\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{ClO}}_4{}^{-}\left(\text{a}\text{q}\right)$

One mole of ${\text{HClO}}_4$ produces one mole of ${\text{H}}^{+}$ ions in the solution, Hence, the concentration of the ${\text{HClO}}_4$ solution and the concentration of ${\text{H}}^{+}$ ions in the solution is equal.

Therefore, the concentration of ${\text{H}}^{+}$ ions in the solution is $0.00562\text{M}$.

The $\text{pH}$ of a solution is represented as,

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$

Where,

• $\left[{\text{H}}^{+}\right]$ represents the concentration of ${\text{H}}^{+}$ ions in the solution.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\text{log}\left(0.00562\text{M}\right)\\ =2.2503\end{array}$

The $\text{pH}$ of the solution is $2.2503$.

The relation between $\text{pH}$ and $\text{pOH}$ of a solution is given as,

$\text{pOH}=\text{14}-\text{pH}$

Substitute the value of $\text{pH}$ in the above equation.

$\begin{array}{c}\text{pOH}=\text{14}-2.2503\\ =11.7497\end{array}$

The $\text{pOH}$ of the solution is $11.7497$.

Interpretation Introduction

(b)

Interpretation:

The $\text{pH}$ and $\text{pOH}$ values for $3.98\times {10}^{-4}\text{M}\text{KOH}$ solution are to be calculated.

Concept Introduction:

The term $\text{pH}$ referred to the power of hydrogen. The $\text{pH}$ of a solution is mathematically equal to the negative logarithm of concentration of ${\text{H}}^{+}$ ion in solution. The $\text{pH}$ of a solution is represented as,

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$

Where,

• $\left[{\text{H}}^{+}\right]$ represents the concentration of ${\text{H}}^{+}$ ions in the solution.

The $\text{pOH}$ of a solution is used to measure the alkalinity of the solution. The $\text{pOH}$ of a solution is mathematically equal to the negative logarithm of concentration of ${\text{OH}}^{-}$ ion in solution. The $\text{pOH}$ of a solution is represented as,

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ represents the concentration of ${\text{OH}}^{-}$ ions in the solution.

Answer to Problem 20CR

The $\text{pH}$ and $\text{pOH}$ values for $3.98\times {10}^{-4}\text{M}\text{KOH}$ solution are $10.5999$ and $3.4001$ respectively.

Explanation of Solution

The concentration of the $\text{KOH}$ solution is $3.98\times {10}^{-4}\text{M}$.

The dissociation of $\text{KOH}$ in aqueous medium is represented as,

$\text{KOH}\left(\text{a}\text{q}\right)\to {\text{K}}^{+}\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

One mole of $\text{KOH}$ produces one mole of ${\text{OH}}^{-}$ ions in the solution, Hence, the concentration of the $\text{KOH}$ solution and the concentration of ${\text{OH}}^{-}$ ions in the solution is equal.

Therefore, the concentration of ${\text{OH}}^{-}$ ions in the solution is $3.98\times {10}^{-4}\text{M}$.

The $\text{pOH}$ of a solution is represented as,

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ represents the concentration of ${\text{OH}}^{-}$ ions in the solution.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}\text{pOH}=-\text{log}\left(3.98\times {10}^{-4}\text{M}\right)\\ =3.4001\end{array}$

The $\text{pOH}$ of the solution is $3.4001$.

The relation between $\text{pH}$ and $\text{pOH}$ of a solution is given as,

$\text{pH}=\text{14}-\text{pOH}$

Substitute the value of $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}=\text{14}-3.4001\\ =10.5999\end{array}$

The $\text{pH}$ of the solution is $10.5999$.

Interpretation Introduction

(c)

Interpretation:

The $\text{pH}$ and $\text{pOH}$ values for $0.078\text{M}{\text{HNO}}_3$ solution are to be calculated.

Concept Introduction:

The term $\text{pH}$ referred to the power of hydrogen. The $\text{pH}$ of a solution is mathematically equal to the negative logarithm of concentration of ${\text{H}}^{+}$ ion in solution. The $\text{pH}$ of a solution is represented as,

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$

Where,

• $\left[{\text{H}}^{+}\right]$ represents the concentration of ${\text{H}}^{+}$ ions in the solution.

The $\text{pOH}$ of a solution is used to measure the alkalinity of the solution. The $\text{pOH}$ of a solution is mathematically equal to the negative logarithm of concentration of ${\text{OH}}^{-}$ ion in solution. The $\text{pOH}$ of a solution is represented as,

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ represents the concentration of ${\text{OH}}^{-}$ ions in the solution.

Answer to Problem 20CR

The $\text{pH}$ and $\text{pOH}$ values for $0.078\text{M}{\text{HNO}}_3$ solution are $1.1079$ and $12.8921$ respectively.

Explanation of Solution

The concentration of the ${\text{HNO}}_3$ solution is $0.078\text{M}$.

The dissociation of ${\text{HNO}}_3$ in aqueous medium is represented as,

${\text{HNO}}_3\left(\text{a}\text{q}\right)\to {\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{NO}}_3{}^{-}\left(\text{a}\text{q}\right)$

One mole of ${\text{HNO}}_3$ produces one mole of ${\text{H}}^{+}$ ions in the solution, Hence, the concentration of the ${\text{HNO}}_3$ solution and the concentration of ${\text{H}}^{+}$ ions in the solution is equal.

Therefore, the concentration of ${\text{H}}^{+}$ ions in the solution is $0.078\text{M}$.

The $\text{pH}$ of a solution is represented as,

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$

Where,

• $\left[{\text{H}}^{+}\right]$ represents the concentration of ${\text{H}}^{+}$ ions in the solution.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\text{log}\left(0.078\text{M}\right)\\ =1.1079\end{array}$

The $\text{pH}$ of the solution is $1.1079$.

The relation between $\text{pH}$ and $\text{pOH}$ of a solution is given as,

$\text{pOH}=\text{14}-\text{pH}$

Substitute the value of $\text{pH}$ in the above equation.

$\begin{array}{c}\text{pOH}=\text{14}-1.1079\\ =12.8921\end{array}$

The $\text{pOH}$ of the solution is $12.8921$.

Interpretation Introduction

(d)

Interpretation:

The $\text{pH}$ and $\text{pOH}$ values for $4.71\times {10}^{-6}\text{M}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ solution are to be calculated.

Concept Introduction:

The term $\text{pH}$ referred to the power of hydrogen. The $\text{pH}$ of a solution is mathematically equal to the negative logarithm of concentration of ${\text{H}}^{+}$ ion in solution. The $\text{pH}$ of a solution is represented as,

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$

Where,

• $\left[{\text{H}}^{+}\right]$ represents the concentration of ${\text{H}}^{+}$ ions in the solution.

The $\text{pOH}$ of a solution is used to measure the alkalinity of the solution. The $\text{pOH}$ of a solution is mathematically equal to the negative logarithm of concentration of ${\text{OH}}^{-}$ ion in solution. The $\text{pOH}$ of a solution is represented as,

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ represents the concentration of ${\text{OH}}^{-}$ ions in the solution.

Answer to Problem 20CR

The $\text{pH}$ and $\text{pOH}$ values for $4.71\times {10}^{-6}\text{M}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ solution are $8.974$ and $5.026$ respectively.

Explanation of Solution

The concentration of the $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ solution is $4.71\times {10}^{-6}\text{M}$.

The dissociation of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ in aqueous medium is represented as,

$\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{a}\text{q}\right)\to {\text{Ca}}^{2+}\left(\text{a}\text{q}\right)+2{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

One mole of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ produces two moles of ${\text{OH}}^{-}$ ions in the solution, Hence, the concentration of ${\text{OH}}^{-}$ ions in the solution is twice as the concentration of the $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ solution. The relation between the number of concentration of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{OH}}^{-}$ in the solution is given as,

${\text{M}}_1=2{\text{M}}_2$

Where,

• ${\text{M}}_1$ represents the concentration of ${\text{OH}}^{-}$ in the solution.
• ${\text{M}}_2$ represents the concentration of the $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ solution.

Substitute the value of ${\text{M}}_1$ and ${\text{M}}_2$ in the above equation.

$\begin{array}{c}{\text{M}}_1=\left(2\right)\left(4.71\times {10}^{-6}\text{M}\right)\\ =9.42\times {10}^{-6}\text{M}\end{array}$

Therefore, the concentration of ${\text{OH}}^{-}$ ions in the solution is $9.42\times {10}^{-6}\text{M}$.

The $\text{pOH}$ of a solution is represented as,

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ represents the concentration of ${\text{OH}}^{-}$ ions in the solution.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}\text{pOH}=-\text{log}\left(9.42\times {10}^{-6}\text{M}\right)\\ =5.026\end{array}$

The $\text{pOH}$ of the solution is $5.026$.

The relation between $\text{pH}$ and $\text{pOH}$ of a solution is given as,

$\text{pH}=\text{14}-\text{pOH}$

Substitute the value of $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}=\text{14}-5.026\\ =8.974\end{array}$

The $\text{pH}$ of the solution is $8.974$.