Chapter 14, Problem 60E

(a)

Interpretation Introduction

Interpretation: The titration of ${\text{H}}_2{\text{NNH}}_2$ with different volumes of ${\text{HNO}}_{\text{3}}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added to it.

Answer to Problem 60E

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{10.74}$.

Explanation of Solution

Explanation

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{10.74}$.

Given:

The concentration of ${\text{H}}_2{\text{NNH}}_2$ is $0.100\text{M}$

The concentration of ${\text{HNO}}_{\text{3}}$ is $0.200\text{M}$.

The volume of ${\text{H}}_2{\text{NNH}}_2$ is $100.0\text{mL}$.

The volume of ${\text{HNO}}_{\text{3}}$ is $0.0\text{mL}$.

The value of $K_{\text{b}}$ of ${\text{H}}_2{\text{NNH}}_2$ is $3.0\times {10}^{-6}$ .

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $100\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{100}\text{mL}=100\times 0.001\text{L}\\ =\text{0}\text{.100}\text{L}\end{array}$

The reaction is represented as,

${\text{H}}_2{\text{NNH}}_2+{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}_2{\text{NNH}}_3{}^{+}+{\text{OH}}^{-}$

Make the ICE table for the reaction between ${\text{H}}_2{\text{NNH}}_2$ and ${\text{H}}_{\text{2}}\text{O}$.

$\begin{array}{cccccccc}& {\text{H}}_2{\text{NNH}}_2& +& {\text{H}}_{\text{2}}\text{O}& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_3{}^{+}& +& \text{OH}-\\ \text{Initial moles:}& 0.100& & & & 0& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.100-x& & & & x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]}\\ \text{3}\text{.0}\times {\text{10}}^{-6}=\frac{\left(x\right)\left(x\right)}{\left(0.100-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.100-x\right)$ is taken as $0.100$.

Simplify the above equation,

$\begin{array}{c}3.0\times {10}^{-6}=\frac{\left(x\right)\left(x\right)}{\left(0.100\right)\text{M}}\\ x=5.4\times {10}^{-4}\text{}\text{M}\end{array}$

It is the concentration of ${\text{OH}}^{-}$.

The $\text{pOH}$ of the solution is calculated by the formula,

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

Where,

• $\left[{\text{OH}}^{-}\right]$ is the concentration of Hydroxide ions.

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(5.4\times {10}^{-4}\right)\\ =3.26\end{array}$

The relationship between $\text{pOH}$ is given as,

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

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

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+\text{3}\text{.26}=14\\ \text{pH}=\underset{\_}{10.74}\end{array}$

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{10.74}$.

(b)

Interpretation Introduction

Interpretation: The titration of ${\text{H}}_2{\text{NNH}}_2$ with different volumes of ${\text{HNO}}_{\text{3}}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $20.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added to it.

Answer to Problem 60E

The value of $\text{pH}$ of solution when $20.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is $\underset{\_}{8.7}$.

Explanation of Solution

Explanation

The value of $\text{pH}$ of solution when $20.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is $\underset{\_}{8.7}$.

Given

The volume of ${\text{HNO}}_{\text{3}}$ is $20.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $20\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{20}\text{mL}=20\times 0.001\text{L}\\ =\text{0}\text{.02}\text{L}\end{array}$

The concentration of any species is given as,

$\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}$ (1)

Rearrange the above equation to obtain the value of number of moles.

$\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}$ (2)

Substitute the value of concentration and volume of ${\text{HNO}}_{\text{3}}$ in equation (4) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.200}\text{M}\times 0.02\text{L}\\ =\text{0}\text{.004}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{H}}_2{\text{NNH}}_2$ and ${\text{HNO}}_{\text{3}}$.

$\begin{array}{cccccccc}& {\text{H}}_2{\text{NNH}}_{\text{2}}& +& {\text{HNO}}_3& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_3{}^{+}& +& {\text{NO}}_3{}^{-}\\ \text{Initial moles:}& 0.01& & 0.004& & 0& & \\ \text{Change}:& -0.004& & -0.004& & +0.004& & \\ \text{Final}\text{moles}:& 0.006& & 0& & 0.004& & \end{array}$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{H}}_2{\text{NNH}}_2+\text{Volume}\text{of}{\text{HNO}}_{\text{3}}\\ =\text{0}\text{.1}\text{L}+0.052\text{L}\\ =\text{0}\text{.12}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{H}}_2{\text{NNH}}_2$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.006\text{moles}}{0.12\text{}\text{L}}\\ =0.05\text{M}\end{array}$

Substitute the value of number of moles of ${\text{H}}_2{\text{NNH}}_3{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.004\text{moles}}{0.12\text{}\text{L}}\\ =0.033\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{H}}_2{\text{NNH}}_2$.

$\begin{array}{cccccccc}& {\text{H}}_2{\text{NNH}}_{\text{2}}& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_3{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.05& & & & 0.033& & \\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.05-x& & & & 0.033+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]}\\ \text{3}\text{.0}\times {\text{10}}^{-6}=\frac{\left(0.033+x\right)\left(x\right)}{\left(0.05-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.05-x\right)$ is taken as $\left(0.05\right)$ and $\left(0.033+x\right)$ is taken as $0.033$.

Simplify the above equation,

$\begin{array}{c}3.0\times {10}^{-6}=\frac{\left(0.033\right)\left(x\right)}{\left(0.05\right)\text{M}}\\ x=4.5\times {10}^{-6}\text{}\text{M}\end{array}$

It is the concentration of ${\text{OH}}^{-}$.

The $\text{pOH}$ of the solution is calculated by the formula,

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(4.5\times {10}^{-6}\right)\\ =5.35\end{array}$

The relationship between $\text{pOH}$ is given as,

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

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

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+\text{5}\text{.35}=14\\ \text{pH}=\underset{\_}{8.7}\end{array}$

The value of $\text{pH}$ of solution when $20.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{8.7}$.

(c)

Interpretation Introduction

Interpretation: The titration of ${\text{H}}_2{\text{NNH}}_2$ with different volumes of ${\text{HNO}}_{\text{3}}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $25.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added to it.

Answer to Problem 60E

The value of $\text{pH}$ of solution when $25.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{8.48}$.

Explanation of Solution

Explanation

The value of $\text{pH}$ of solution when $25.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{8.48}$.

Given

The volume of ${\text{HNO}}_{\text{3}}$ is $25.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $25\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{25}\text{mL}=25\times 0.001\text{L}\\ =\text{0}\text{.025}\text{L}\end{array}$

Substitute the value of concentration and volume of ${\text{HNO}}_{\text{3}}$ in equation (4) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.200}\text{M}\times 0.025\text{L}\\ =\text{0}\text{.005}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{H}}_2{\text{NNH}}_2$ and ${\text{HNO}}_{\text{3}}$.

$\begin{array}{cccccccc}& {\text{H}}_2{\text{NNH}}_{\text{2}}& +& {\text{HNO}}_3& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_3{}^{+}& +& {\text{NO}}_3{}^{-}\\ \text{Initial moles:}& 0.01& & 0.005& & 0& & \\ \text{Change}:& -0.005& & -0.005& & +0.005& & \\ \text{Final}\text{moles}:& 0.005& & 0& & 0.005& & \end{array}$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{H}}_2{\text{NNH}}_2+\text{Volume}\text{of}{\text{HNO}}_{\text{3}}\\ =\text{0}\text{.1}\text{L}+0.025\text{L}\\ =\text{0}\text{.125}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{H}}_2{\text{NNH}}_2$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.005\text{moles}}{0.125\text{}\text{L}}\\ =0.04\text{M}\end{array}$

Substitute the value of number of moles of ${\text{H}}_2{\text{NNH}}_3{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.005\text{moles}}{0.125\text{}\text{L}}\\ =0.04\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{H}}_2{\text{NNH}}_2$.

$\begin{array}{cccccccc}& {\text{H}}_2{\text{NNH}}_{\text{2}}& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_3{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.04& & & & 0.04& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.04-x& & & & 0.04+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]}\\ \text{3}\text{.0}\times {\text{10}}^{-6}=\frac{\left(0.04+x\right)\left(x\right)}{\left(0.04-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.04-x\right)$ is taken as $\left(0.04\right)$ and $\left(0.04+x\right)$ is taken as $0.04$.

Simplify the above equation,

$\begin{array}{c}3.0\times {10}^{-6}=\frac{\left(0.04\right)\left(x\right)}{\left(0.04\right)\text{M}}\\ x=3.0\times {10}^{-6}\text{}\text{M}\end{array}$

It is the concentration of ${\text{OH}}^{-}$.

The $\text{pOH}$ of the solution is calculated by the formula,

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(3.0\times {10}^{-6}\right)\\ =5.52\end{array}$

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

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+\text{5}\text{.52}=14\\ \text{pH}=\underset{\_}{8.48}\end{array}$

The value of $\text{pH}$ of solution when $25.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{8.48}$.

(d)

Interpretation Introduction

Interpretation: The titration of ${\text{H}}_2{\text{NNH}}_2$ with different volumes of ${\text{HNO}}_{\text{3}}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $40.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added to it.

Answer to Problem 60E

The value of $\text{pH}$ of solution when $40.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{7.87}$.

Explanation of Solution

Explanation

The value of $\text{pH}$ of solution when $40.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{7.87}$.

Given

The volume of ${\text{HNO}}_{\text{3}}$ is $40.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $40\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{40}\text{mL}=40\times 0.001\text{L}\\ =\text{0}\text{.04}\text{L}\end{array}$

Substitute the value of concentration and volume of ${\text{HNO}}_{\text{3}}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.200}\text{M}\times 0.04\text{L}\\ =\text{0}\text{.008}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{H}}_2{\text{NNH}}_2$ and ${\text{HNO}}_{\text{3}}$.

$\begin{array}{cccccccc}& {\text{H}}_2{\text{NNH}}_{\text{2}}& +& {\text{HNO}}_3& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_3{}^{+}& +& {\text{NO}}_3{}^{-}\\ \text{Initial moles:}& 0.01& & 0.008& & 0& & \\ \text{Change}:& -0.008& & -0.008& & +0.008& & \\ \text{Final}\text{moles}:& 0.002& & 0.002& & 0.008& & \end{array}$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{H}}_2{\text{NNH}}_2+\text{Volume}\text{of}{\text{HNO}}_{\text{3}}\\ =\text{0}\text{.1}\text{L}+0.04\text{L}\\ =\text{0}\text{.14}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{H}}_2{\text{NNH}}_2$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.002\text{moles}}{0.14\text{}\text{L}}\\ =0.014\text{M}\end{array}$

Substitute the value of number of moles of ${\text{H}}_2{\text{NNH}}_3{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.008\text{moles}}{0.14\text{}\text{L}}\\ =0.057\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{H}}_2{\text{NNH}}_2$.

$\begin{array}{cccccccc}& {\text{H}}_2{\text{NNH}}_{\text{2}}& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_3{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.014& & & & 0.0571& & \\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.014-x& & & & 0.0571+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]}\\ \text{3}\text{.0}\times {\text{10}}^{-6}=\frac{\left(0.0571+x\right)\left(x\right)}{\left(0.014-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.014-x\right)$ is taken as $\left(0.014\right)$ and $\left(0.0571+x\right)$ is taken as $0.0571$.

Simplify the above equation,

$\begin{array}{c}3.0\times {10}^{-6}=\frac{\left(0.0571\right)\left(x\right)}{\left(0.014\right)\text{M}}\\ x=0.74\times {10}^{-6}\text{}\text{M}\end{array}$

It is the concentration of ${\text{OH}}^{-}$.

The $\text{pOH}$ of the solution is calculated by the formula,

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.74\times {10}^{-6}\right)\\ =6.13\end{array}$

The relationship between $\text{pOH}$ is given as,

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

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

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+\text{6}\text{.13}=14\\ \text{pH}=\underset{\_}{7.87}\end{array}$

The value of $\text{pH}$ of solution when $40.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{7.87}$.

(e)

Interpretation Introduction

Interpretation: The titration of ${\text{H}}_2{\text{NNH}}_2$ with different volumes of ${\text{HNO}}_{\text{3}}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $50.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added to it.

Answer to Problem 60E

The value of $\text{pH}$ of solution when $50.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{4.85}$.

Explanation of Solution

Explanation

The value of $\text{pH}$ of solution when $50.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{4.85}$.

Given

The volume of ${\text{HNO}}_{\text{3}}$ is $50.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $50\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{50}\text{mL}=50\times 0.001\text{L}\\ =\text{0}\text{.05}\text{L}\end{array}$

Substitute the value of concentration and volume of ${\text{HNO}}_{\text{3}}$ in equation (2) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.200}\text{M}\times 0.05\text{L}\\ =\text{0}\text{.01}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{H}}_2{\text{NNH}}_2$ and ${\text{HNO}}_{\text{3}}$.

$\begin{array}{cccccccc}& {\text{H}}_2{\text{NNH}}_{\text{2}}& +& {\text{HNO}}_3& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_3{}^{+}& +& {\text{NO}}_3{}^{-}\\ \text{Initial moles:}& 0.01& & 0.01& & 0& & \\ \text{Change}:& -0.01& & -0.01& & +0.01& & \\ \text{Final}\text{moles}:& 0& & 0& & 0.01& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{H}}_2{\text{NNH}}_2+\text{Volume}\text{of}{\text{HNO}}_{\text{3}}\\ =\text{0}\text{.1}\text{L}+0.05\text{L}\\ =\text{0}\text{.15}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{H}}_2{\text{NNH}}_3{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.01\text{moles}}{0.15\text{}\text{L}}\\ =0.067\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{H}}_2{\text{NNH}}_3{}^{+}$.

$\begin{array}{ccccc}& {\text{H}}_2{\text{NNH}}_3{}^{+}& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_{\text{2}}& {\text{H}}^{+}\\ \text{Initial}\text{(M)}:& 0.067& & 0& 0\\ \text{Change}\left(\text{M}\right):& -x& & x& x\\ \text{Equilibrium}\left(\text{M}\right):& 0.067-x& & x& x\end{array}$

The conjugate acid-base pair are related to each other through a formula as shown below.

$K_{\text{w}}=K_{\text{a}}\times K_{\text{b}}$

Where,

• $K_{\text{a}}$ is the dissociation constant of an acid.

• $K_{\text{b}}$ is the dissociation constant of a base.

• $K_{\text{w}}$ is the ionic product of water $\left({10}^{-14}\right)$.

Rearrange the equation to obtain the value of $K_{\text{a}}$ as,

$K_{\text{a}}=\frac{K_{\text{w}}}{K_{\text{b}}}$

Substitute the value of $K_{\text{w}}$ and $K_{\text{b}}$ in the above equation as,

$\begin{array}{c}{K}_{\text{a}}=\frac{K_{\text{w}}}{K_{\text{b}}}\\ =\frac{{10}^{-14}}{3\times {10}^{-6}}\\ =3.3\times {10}^{-9}\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{H}}_2{\text{NNH}}_{\text{2}}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]}\\ 3.3\times {\text{10}}^{-\text{9}}=\frac{\left(x\right)\left(x\right)}{\left(0.067-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{a}}$ is very small, hence, $\left(0.067-x\right)$ is taken as $0.067$.

Simplify the above equation,

$\begin{array}{c}3.3\times {\text{10}}^{-\text{9}}=\frac{\left(x\right)\left(x\right)}{\left(0.067\right)\text{M}}\\ x=1.41\times {10}^{-5}\text{M}\end{array}$

The $\text{pH}$ of the solution is calculated by the formula,

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

Where,

• $\left[{\text{H}}^{+}\right]$ is the concentration of Hydrogen ions.

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

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(1.41\times {10}^{-5}\right)\\ =\underset{\_}{4.85}\end{array}$

The value of $\text{pH}$ of solution when $50.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is. $\underset{\_}{4.85}$.

(f)

Interpretation Introduction

Interpretation: The titration of ${\text{H}}_2{\text{NNH}}_2$ with different volumes of ${\text{HNO}}_{\text{3}}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $100.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added to it.

Answer to Problem 60E

The value of $\text{pH}$ of solution when $100.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is $\underset{\_}{1.3}$.

Explanation of Solution

Explanation

The value of $\text{pH}$ of solution when $100.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is $\underset{\_}{1.3}$.

Given

The volume of ${\text{HNO}}_{\text{3}}$ is $100.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $100.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{100}\text{mL}=100\times 0.001\text{L}\\ =\text{0}\text{.1}\text{L}\end{array}$

Substitute the value of concentration and volume of ${\text{HNO}}_{\text{3}}$ in equation (2) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.200}\text{M}\times 0.1\text{L}\\ =\text{0}\text{.02}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{H}}_2{\text{NNH}}_2$ and ${\text{HNO}}_{\text{3}}$.

$\begin{array}{cccccccc}& {\text{H}}_2{\text{NNH}}_{\text{2}}& +& {\text{HNO}}_3& \stackrel{}{\rightleftharpoons }& {\text{H}}_2{\text{NNH}}_3{}^{+}& +& {\text{NO}}_3{}^{-}\\ \text{Initial moles:}& 0.01& & 0.02& & 0& & \\ \text{Change}:& -0.01& & -0.01& & +0.01& & \\ \text{Final}\text{moles}:& 0& & 0.01& & 0.01& & \end{array}$

There is excess of Hydrogen ions in the solution

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{H}}_2{\text{NNH}}_2+\text{Volume}\text{of}{\text{HNO}}_{\text{3}}\\ =\text{0}\text{.1}\text{L}+0.1\text{L}\\ =\text{0}\text{.2}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{H}}^{+}$ and final volume of solution in equation (1)

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.01\text{moles}}{0.2\text{}\text{L}}\\ =0.05\text{M}\end{array}$

The $\text{pH}$ of the solution is calculated by the formula,

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

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

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.05\right)\\ =\underset{\_}{1.3}\end{array}$

The value of $\text{pH}$ of solution when $100.0\text{}\text{mL}$ ${\text{HNO}}_{\text{3}}$ has been added is $\underset{\_}{1.3}$.