Chapter 15, Problem 64AE

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ of solution formed $\text{50}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ HCl}$ is titrated with $\text{0}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ NaOH}$ has to be calculated.

Concept Introduction:

The formula to calculate molar mass from number of moles is given as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

The expression to evaluate $\text{pH}$ is as follows:

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

Explanation of Solution

Since in $\text{1000 mL}$ moles of $\text{HCl}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{HCl}$ in $\text{50}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{50}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.01\text{ mol}\end{array}$

The formula to evaluate molarity is given as follows:

  $M=\frac{n}{V}$        (1)

Substitute $0.01\text{ mol}$ for $n$ and $\text{50 mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{0.01\text{ mol}}{\text{50 mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.2\text{ M}\end{array}$

Thus concentration of ${\text{H}}^{+}$ is $0.2\text{ M}$.

The expression to evaluate $\text{pH}$ is as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$        (2)

Substitute $0.2\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (2) to calculate $\text{pH}$.

  $\begin{array}{c}\text{pH}=-\log \left(0.2\text{ M}\right)\\ =0.6989\end{array}$

Hence, $\text{pH}$ of solution is 0.6989.

(b)

Interpretation Introduction

Interpretation:

Whether solution formed $\text{50}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ HCl}$ is titrated with $\text{10}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since in $\text{1000 mL}$ moles of $\text{HCl}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{HCl}$ in $\text{50}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{50}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.01\text{ mol}\end{array}$

Since in $\text{1000 mL}$ moles of $\text{NaOH}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{NaOH}$ in $\text{10}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}=\left(\text{10}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.002\text{ mol}\end{array}$

The balanced equation is given as follows:

  $\text{HCl}+\text{NaOH}\to \text{NaCl}+{\text{H}}_{\text{2}}\text{O}$        (3)

As per balanced equation (3) since $\text{1 mol NaOH}$ reacts with $\text{1 mol HCl}$ so moles of $\text{HCl}$ that should react with $0.002\text{ mol NaOH}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(0.002\text{ mol NaOH}\right)\left(\frac{\text{1 mol HCl}}{\text{1 mol NaOH}}\right)\\ =0.002\text{ mol HCl}\end{array}$

Since moles of $\text{HCl}$ that available to react are $0.01\text{ mol}$ while that should react as per reaction is $0.002\text{ mol}$ thus $\text{NaOH}$ is limiting reactant and hence excess concentration of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}\text{Excess HCl}=0.01\text{ mol}-0.002\text{ mol}\\ =\text{0}\text{.008 mol}\end{array}$

Total volume is calculated as follows:

  $\begin{array}{c}\text{Total volume}=50\text{ mL}+10\text{ mL}\\ =60\text{ mL}\end{array}$

Substitute $\text{0}\text{.008 mol}$ for $n$ and $60\text{ mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{\text{0}\text{.008 mol}}{60\text{ mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.1333\text{ M}\end{array}$

Thus concentration of ${\text{H}}^{+}$ is $\text{0}\text{.008 }M$.

Substitute $0.1333\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (2) to calculate $\text{pH}$.

  $\begin{array}{c}\text{pH}=-\log \left(0.1333\text{ M}\right)\\ =0.875\end{array}$

Hence, $\text{pH}$ of solution is 0.875.

(c)

Interpretation Introduction

Interpretation:

Whether solution formed $\text{50}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ HCl}$ is titrated with $\text{25}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since in $\text{1000 mL}$ moles of $\text{HCl}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{HCl}$ in $\text{50}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{50}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.01\text{ mol}\end{array}$

Since in $\text{1000 mL}$ moles of $\text{NaOH}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{NaOH}$ in $\text{25}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}=\left(\text{25}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.005\text{ mol}\end{array}$

As per balanced equation (3) since $\text{1 mol NaOH}$ reacts with $\text{1 mol HCl}$ so moles of $\text{HCl}$ that should react with $0.005\text{ mol NaOH}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(0.005\text{ mol NaOH}\right)\left(\frac{\text{1 mol HCl}}{\text{1 mol NaOH}}\right)\\ =0.005\text{ mol HCl}\end{array}$

Since moles of $\text{HCl}$ that are available to react are $0.01\text{ mol}$ while that should react as per reaction is $0.005\text{ mol}$ thus $\text{NaOH}$ is limiting reactant and hence excess concentration of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}\text{Excess HCl}=0.01\text{ mol}-0.005\text{ mol}\\ =\text{0}\text{.005 mol}\end{array}$

Total volume is calculated as follows:

  $\begin{array}{c}\text{Total volume}=50\text{ mL}+\text{25}\text{.0 mL}\\ =75\text{ mL}\end{array}$

Substitute $\text{0}\text{.005 mol}$ for $n$ and $75\text{ mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{\text{0}\text{.005 mol}}{75\text{ mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.0667\text{ M}\end{array}$

Substitute $0.0667\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (2) to calculate $\text{pH}$.

  $\begin{array}{c}\text{pH}=-\log \left(0.0667\text{ M}\right)\\ =1.176\end{array}$

Hence, $\text{pH}$ of solution is 1.176.

(d)

Interpretation Introduction

Interpretation:

Whether solution formed $\text{50}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ HCl}$ is titrated with $\text{49}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since in $\text{1000 mL}$ moles of $\text{HCl}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{HCl}$ in $\text{50}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{50}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.01\text{ mol}\end{array}$

Since in $\text{1000 mL}$ moles of $\text{NaOH}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{NaOH}$ in $\text{49}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}=\left(\text{49}\text{.0 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.0098\text{ mol}\end{array}$

As per balanced equation (3) since $\text{1 mol NaOH}$ reacts with $\text{1 mol HCl}$ so moles of $\text{HCl}$ that should react with $0.0098\text{ mol NaOH}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(0.0098\text{ mol NaOH}\right)\left(\frac{\text{1 mol HCl}}{\text{1 mol NaOH}}\right)\\ =0.0098\text{ mol HCl}\end{array}$

Since moles of $\text{HCl}$ that available to react are $0.01\text{ mol}$ while that should react as per reaction is $0.0098\text{ mol}$ thus $\text{NaOH}$ is limiting reactant and hence excess concentration of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}\text{Excess HCl}=0.01\text{ mol}-0.0098\text{ mol}\\ =\text{0}\text{.0002 mol}\end{array}$

Total volume is calculated as follows:

  $\begin{array}{c}\text{Total volume}=50\text{ mL}+\text{49}\text{.0 mL}\\ =99\text{ mL}\end{array}$

Substitute $\text{0}\text{.0002 mol}$ for $n$ and $99\text{ mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{\text{0}\text{.0002 mol}}{99\text{ mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.0020\text{ M}\end{array}$

Thus concentration of ${\text{H}}^{+}$ is $0.0020\text{ M}$.

Substitute $0.0020\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (1) to calculate $\text{pH}$.

  $\begin{array}{c}\text{pH}=-\log \left(0.0020\text{ M}\right)\\ =2.6989\end{array}$

Hence, $\text{pH}$ of solution is 2.6989.

(e)

Interpretation Introduction

Interpretation:

Whether solution formed $\text{50}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ HCl}$ is titrated with $\text{49}\text{.90 mL}$ of $\text{0}\text{.2000 }M\text{ NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since in $\text{1000 mL}$ moles of $\text{HCl}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{HCl}$ in $\text{50}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{50}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.01\text{ mol}\end{array}$

Since in $\text{1000 mL}$ moles of $\text{NaOH}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{NaOH}$ in $\text{49}\text{.90 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}=\left(\text{49}\text{.90 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.00998\text{ mol}\end{array}$

As per balanced equation (3) since $\text{1 mol NaOH}$ reacts with $\text{1 mol HCl}$ so moles of $\text{HCl}$ that should react with $0.00998\text{ mol NaOH}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(0.00998\text{ mol NaOH}\right)\left(\frac{\text{1 mol HCl}}{\text{1 mol NaOH}}\right)\\ =0.00998\text{ mol HCl}\end{array}$

Since moles of $\text{HCl}$ that available to react are $0.01\text{ mol}$ while that should react as per reaction is $0.00998\text{ mol}$ thus $\text{HCl}$ is limiting reactant and hence excess concentration of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}\text{Excess HCl}=0.01\text{ mol}-0.00998\text{ mol}\\ =\text{0}\text{.00002 mol}\end{array}$

Total volume is calculated as follows:

  $\begin{array}{c}\text{Total volume}=50\text{ mL}+\text{49}\text{.90 mL}\\ =99.90\text{ mL}\end{array}$

Substitute $\text{0}\text{.00002 mol}$ for $n$ and $99.90\text{ mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{\text{0}\text{.00002 mol}}{99.90\text{ mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =2\times {10}^{-4}\text{ M}\end{array}$

Substitute $2\times {10}^{-4}\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (2) to calculate $\text{pH}$.

  $\begin{array}{c}\text{pH}=-\log \left(2\times {10}^{-4}\text{ M}\right)\\ =3.6985\end{array}$

Hence, $\text{pH}$ of solution is 3.6985.

(f)

Interpretation Introduction

Interpretation:

Whether solution formed $\text{50}\text{.0 mL}$ of $\text{0}\text{.2000 }M\text{ HCl}$ is titrated with $\text{49}\text{.99 mL}$ of $\text{0}\text{.2000 }M\text{ NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since in $\text{1000 mL}$ moles of $\text{HCl}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{HCl}$ in $\text{50}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{50}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.01\text{ mol}\end{array}$

Since in $\text{1000 mL}$ moles of $\text{NaOH}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{NaOH}$ in $\text{49}\text{.99 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}=\left(\text{49}\text{.99 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.009998\text{ mol}\end{array}$

As per balanced equation (3) since $\text{1 mol NaOH}$ reacts with $\text{1 mol HCl}$ so moles of $\text{HCl}$ that should react with $0.00998\text{ mol NaOH}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(0.009998\text{ mol NaOH}\right)\left(\frac{\text{1 mol HCl}}{\text{1 mol NaOH}}\right)\\ =0.009998\text{ mol HCl}\end{array}$

Since moles of $\text{HCl}$ that available to react are $0.01\text{ mol}$ while that should react as per reaction is $0.009998\text{ mol}$ thus $\text{HCl}$ is limiting reactant and hence excess concentration of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}\text{Excess HCl}=0.01\text{ mol}-0.009998\text{ mol}\\ =\text{0}\text{.000002 mol}\end{array}$

Total volume is calculated as follows:

  $\begin{array}{c}\text{Total volume}=50\text{ mL}+\text{49}\text{.99 mL}\\ =99.99\text{ mL}\end{array}$

Substitute $\text{0}\text{.000002 mol}$ for $n$ and $99.99\text{ mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{\text{0}\text{.000002 mol}}{99.99\text{ mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =2\times {10}^{-5}\text{ M}\end{array}$

Substitute $2\times {10}^{-5}\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (2) to calculate $\text{pH}$.

  $\begin{array}{c}\text{pH}=-\log \left(2\times {10}^{-5}\text{ M}\right)\\ =4.6989\end{array}$

Hence, $\text{pH}$ of solution is 4.6989.

(g)

Interpretation Introduction

Interpretation:

Whether solution formed $\text{50}\text{.00 mL}$ of $\text{0}\text{.2000 }M\text{ HCl}$ is titrated with $\text{50}\text{.00 mL}$ of $\text{0}\text{.2000 }M\text{ NaOH}$ has to be calculated and plot of $\text{pH}$ against volume of $\text{NaOH}$ has to be drawn.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since in $\text{1000 mL}$ moles of $\text{HCl}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{HCl}$ in $\text{50}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{50}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.01\text{ mol}\end{array}$

Since in $\text{1000 mL}$ moles of $\text{NaOH}$ present is $\text{0}\text{.2000 mol}$ so, moles of $\text{NaOH}$ in $\text{50}\text{.00 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}=\left(\text{50}\text{.00 mL}\right)\left(\frac{\text{0}\text{.2000 mol}}{\text{1000 mL}}\right)\\ =0.01\text{ mol}\end{array}$

As per balanced equation (3) since $\text{1 mol NaOH}$ reacts with $\text{1 mol HCl}$ so $0.01\text{ mol NaOH}$ will be completely neutralized by $0.01\text{ mol HCl}$. So solution would be neutral and hence $\text{pH}$ will be 7.

The table for plot of $\text{pH}$ against volume of $\text{NaOH}$ is given as follows:

$\begin{array}{cc}\text{Volume of NaOH}& \text{pH}\\ 0.000\text{ mL}& 0.6989\\ 10.00\text{ mL}& 0.875\\ 25.00\text{ mL}& 1.176\\ 49.00\text{ mL}& 2.6989\\ 49.90\text{ mL}& 3.6985\\ 49.99\text{ mL}& 4.6989\\ \text{50}\text{.00 mL}& 7\end{array}$

Plot of $\text{pH}$ against volume of $\text{NaOH}$ is drawn as follows: