Chapter 13, Problem 147QP

When $10.0\text{ mL of a }0.10 M\text{ HCl}$ solution is diluted to $\text{1}\text{.00}\text{L},$ what is the pH of the new solution?

$\begin{array}{l}\text{A}.\text{pH}=-1.00\\ \text{B}\text{.}\text{pH}=1.00\\ \text{C}\text{.}\text{pH}=2.00\\ \text{D}\text{.}\text{pH}=3.00\\ \text{E}\text{. pH}=4.00\end{array}$

Which of the other pH values could you obtain by diluting a $0.10 M \text{HCl}$ solution, and how would you prepare each solution with the desired pH?

Interpretation Introduction

Interpretation:

The given statement that $0.1M$ $\text{HCl}$ solution which is diluted in $1\text{ L}$ has $\text{pH}$ value $-1$ or not is to be identified. If not, the concentration of the solution at which this $\text{pH}$ value exists is to be calculated.

Explanation of Solution

It is given that $10\text{ mL}$ of $0.1M$ $\text{HCl}$ solution is diluted in $1\text{ L}$ solution.

$\begin{array}{l}1\text{ L}=1\text{ L}\times \frac{1000\text{ mL}}{1\text{ L}}\\ 1\text{ L}=1000\text{ mL}\end{array}$

It is known that,

$M_1{V}_1=M_2{V}_2$ ….(1)

Substitute $M_1$ , $V_1$ , $M_2$ and $V_2$ as $0.1M$ , $10\text{ mL}$ , $M_2$ and $1000\text{ mL}$ , respectively in equation (1).

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ 0.1M\times 10\text{ mL}=M_2\times 1000\text{ mL}\\ M_2=\frac{0.1M\times 10\text{ mL}}{1000\text{ mL}}\\ M_2={10}^{-3}M\end{array}$

The ${\text{H}}_3{\text{O}}^{+}$ of the new solution is $1\times {10}^{-3}M$ . The $\text{pH}$ is the negative ${\log }_{10}$ of the hydronium ion concentration. Therefore,

$\begin{array}{l}\text{pH}=\left(-{\log }_{10}\left[{\text{H}}_3{\text{O}}^{+}\right]\right)\\ \text{pH}=\left(-{\log }_{10}1\times {10}^{-3}M\right)\\ \text{pH}=3\end{array}$

The $\text{pH}$ of new solution is $3$ . The $\text{pH}$ given is $-1$ . Therefore, the given statement is an incorrect statement.

The $\text{pH}$ given is $-1$ and the hydronium ion concentration at this $\text{pH}$ value is as follows:

$\begin{array}{c}\left(-{\log }_{10}\left[{\text{H}}_3{\text{O}}^{+}\right]\right)=\text{pH}\\ \left(-{\log }_{10}\left[{\text{H}}_3{\text{O}}^{+}\right]\right)=\left(-1\right)\\ \left[{\text{H}}_3{\text{O}}^{+}\right]={10}^1\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=10M\end{array}$

Now, $\text{pH}$ of $-1$ cannot be obtained from $0.1M$ solution as diluting will increase the $\text{pH}$ and hence the resulting concentration will be lower than $10M$ .

Interpretation Introduction

Interpretation:

The given statement that $0.1M$ $\text{HCl}$ solution which is diluted in $1\text{ L}$ has $\text{pH}$ value $1$ or not is to be identified. If not, the concentration of the solution at which this $\text{pH}$ value exists is to be calculated.

Explanation of Solution

The $\text{pH}$ value of the given solution is $3$ as calculated in sub-part (a). Therefore, the given statement is an incorrect statement.

The $\text{pH}$ given is $1$ and the hydronium ion concentration at this $\text{pH}$ value is as follows:

$\begin{array}{c}\left(-{\log }_{10}\left[{\text{H}}_3{\text{O}}^{+}\right]\right)=\text{pH}\\ \left(-{\log }_{10}\left[{\text{H}}_3{\text{O}}^{+}\right]\right)=1\\ \left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-1}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=0.1M\end{array}$

The given solution is already $0.1M$ and hence no dilution is required.

Interpretation Introduction

Interpretation:

The given statement that $0.1M$ $\text{HCl}$ solution which is diluted in $1\text{ L}$ has $\text{pH}$ value $2$ or not is to be identified. If not, the concentration of the solution at which this $\text{pH}$ value exists is to be calculated.

Explanation of Solution

The $\text{pH}$ value of the given solution is $3$ as calculated in sub-part (a). Therefore, the given statement is an incorrect statement.

The $\text{pH}$ given is $2$ and the hydronium ion concentration at this $\text{pH}$ value is as follows:

$\begin{array}{c}\left(-{\log }_{10}\left[{\text{H}}_3{\text{O}}^{+}\right]\right)=\text{pH}\\ \left(-{\log }_{10}\left[{\text{H}}_3{\text{O}}^{+}\right]\right)=2\\ \left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-2}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=0.01M\end{array}$

Substitute $M_1$ , $V_1$ , $M_2$ and $V_2$ as $0.1M$ , $10\text{ mL}$ , $0.01M$ and $V_2$ , respectively in equation (1).

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ 0.1M\times 10\text{ mL}=0.01M\times V_2\\ V_2=\frac{0.1M\times 10\text{ mL}}{0.01M}\\ V_2=100\text{ mL}\end{array}$

The solution needs to be diluted to $100\text{ mL}$ to get the desired $\text{pH}$ .

Interpretation Introduction

Interpretation:

The given statement that $0.1M$ $\text{HCl}$ solution which is diluted in $1\text{ L}$ has $\text{pH}$ value $3$ or not is to be identified. If not, the concentration of the solution at which this $\text{pH}$ value exists is to be calculated.

Explanation of Solution

The $\text{pH}$ value of the given solution is $3$ as calculated in sub-part (a). Therefore, the given statement is a correct statement.

Interpretation Introduction

Interpretation:

The given statement that $0.1M$ $\text{HCl}$ solution which is diluted in $1\text{ L}$ has $\text{pH}$ value $4$ or not is to be identified. If not, the concentration of the solution at which this $\text{pH}$ value exists is to be calculated.

Explanation of Solution

The $\text{pH}$ value of the given solution is $3$ as calculated in sub-part (a). Therefore, the given statement is an incorrect statement.

The $\text{pH}$ given is $2$ and the hydronium ion concentration at this $\text{pH}$ value is as follows:

$\begin{array}{c}\left(-{\log }_{10}\left[{\text{H}}_3{\text{O}}^{+}\right]\right)=\text{pH}\\ \left(-{\log }_{10}\left[{\text{H}}_3{\text{O}}^{+}\right]\right)=4\\ \left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-4}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=0.0001M\end{array}$

Substitute $M_1$ , $V_1$ , $M_2$ and $V_2$ as $0.1M$ , $10\text{ mL}$ , $0.0001M$ and $V_2$ , respectively in equation (1).

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ 0.1M\times 10\text{ mL}=0.0001M\times V_2\\ V_2=\frac{0.1M\times 10\text{ mL}}{0.0001M}\\ V_2=10000\text{ mL}\end{array}$

The solution needs to be diluted to $10000\text{ mL}$ or $10\text{ L}$ to get the desired $\text{pH}$ .