Chapter 16, Problem 16.109QP

Interpretation Introduction

Interpretation: The $\text{pH}$ at ${\text{25}}^{\text{o}}\text{C}$ of $\text{1}\text{.00}\text{L}$ of a buffer that is $\text{0}\text{.120}\text{M}{\text{HNO}}_{\text{2}}$ and $\text{0}\text{.150}\text{M}{\text{NaNO}}_{\text{2}}$ before and after the addition of $\text{1}\text{.00}\text{mL}$ of a $\text{12}\text{.0}\text{M}\text{HCl}$ is to be calculated.

Concept introduction: The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation,

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

Where,

• $\text{p}{K}_{\text{a}}$ is the acid dissociation constant.

The molarity of the solution is calculated by the formula,

$\begin{array}{c}\text{Molarity}=\frac{\text{Amount}\text{of}\text{solute}\left(\text{in}\text{moles}\right)}{\text{Volume}}\\ \text{Moles}=\text{Molarity}\times \text{Volume}\end{array}$

To determine: The $\text{pH}$ at ${\text{25}}^{\text{o}}\text{C}$ of $\text{1}\text{.00}\text{L}$ of a buffer that is $\text{0}\text{.120}\text{M}{\text{HNO}}_{\text{2}}$ and $\text{0}\text{.150}\text{M}{\text{NaNO}}_{\text{2}}$ before and after the addition of $\text{1}\text{.00}\text{mL}$ of a $\text{12}\text{.0}\text{M}\text{HCl}$.

Answer to Problem 16.109QP

The $\text{pH}$ at ${\text{25}}^{\text{o}}\text{C}$ of $\text{1}\text{.00}\text{L}$ of a buffer that is $\text{0}\text{.120}\text{M}{\text{HNO}}_{\text{2}}$ and $\text{0}\text{.150}\text{M}{\text{NaNO}}_{\text{2}}$ before and after the addition of $\text{1}\text{.00}\text{mL}$ of a $\text{12}\text{.0}\text{M}\text{HCl}$ is $\underset{\_}{3.42}$.

Explanation of Solution

Given

The molarity of ${\text{HNO}}_2$ is $\text{0}\text{.120}\text{M}$.

The molarity of ${\text{NaNO}}_{\text{2}}$ is $\text{0}\text{.150}\text{M}$.

The molarity of $\text{HCl}$ is $\text{12}\text{.0}\text{M}$.

The volume of ${\text{HNO}}_2$ and ${\text{NaNO}}_{\text{2}}$ is $\text{1}\text{.00}\text{L}$.

The volume of $\text{HCl}$ solution added is $\text{1}\text{.00}\text{mL}$.

The molarity of the solution is calculated by the formula,

$\begin{array}{c}\text{Molarity}=\frac{\text{Amount}\text{of}\text{solute}\left(\text{in}\text{moles}\right)}{\text{Volume}}\\ \text{Moles}=\text{Molarity}\times \text{Volume}\end{array}$

Substitute the values of molarity and volume

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{NaNO}}_2=\text{Molarity}\times \text{Volume}\\ =\text{0}\text{.150}\text{M}\times \text{1}\text{.00}\text{L}\\ =0.150\text{mol}\end{array}$

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{HNO}}_2=\text{Molarity}\times \text{Volume}\\ =\text{0}\text{.120}\text{M}\times \text{1}\text{.00}\text{L}\\ =\text{0}\text{.120}\text{mol}\end{array}$

The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation,

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)\\ \text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{NaNO}}_{\text{2}}\right]}{\left[{\text{HNO}}_2\right]}\end{array}$

The $\text{p}{K}_{\text{a}}$ value of ${\text{HNO}}_{\text{2}}$ is $3.40$.

Substitute the values in the above equation.

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{NaNO}}_{\text{2}}\right]}{\left[{\text{HNO}}_2\right]}\\ \text{pH}=3.40+\log \frac{\left[0.150\right]}{\left[0.120\right]}\\ \text{pH}=3.40+\log \left[0.150\right]-\log \left[0.120\right]\\ \text{pH}=3.50\end{array}$

With the addition of $1.00\text{mL}$ of $12.0\text{M}\text{HCl}$ :

$\begin{array}{c}\text{Moles}\text{of}{\text{H}}^{\text{+}}\text{added}=\text{Molarity}\times \text{Volume}\\ =12\text{mol}/\text{L}\times 1\times {10}^{-3}\\ =12\times {10}^{-3}\text{moles}\\ =\text{0}\text{.012}\text{moles}/\text{mL}\end{array}$

So, $0.012\text{moles}$ of $\text{HCl}$ will convert $0.012\text{moles}$ of ${\text{NaNO}}_{\text{2}}$ into $0.012\text{moles}$ of ${\text{HNO}}_{\text{2}}$.

So, Final number of moles of ${\text{HNO}}_{\text{2}}$ $\begin{array}{l}=0.120+0.012\\ =0.132\end{array}$

Final number of moles of ${\text{NaNO}}_{\text{2}}$ $\begin{array}{l}=0.150-0.012\\ =0.138\end{array}$

$\begin{array}{c}\text{Total}\text{volume}=1+\frac{1}{1000}\\ =1.001\text{L}\end{array}$

The molarity of the solution is calculated by the formula,

$\text{Molarity}=\frac{\text{Amount}\text{of}\text{solute}\left(\text{in}\text{moles}\right)}{\text{Volume}}$

Substitute the values of molarity and volume.

$\begin{array}{c}\text{Molarity}\text{of}{\text{NaNO}}_2\left(\text{Salt}\right)=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}}\\ =\frac{0.138\text{moles}}{1.001\text{L}}\\ =0.138\text{M}\end{array}$

$\begin{array}{c}\text{Molarity}\text{of}{\text{HNO}}_2\left(\text{acid}\right)=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}}\\ =\frac{0.132\text{moles}}{1.001\text{L}}\\ =0.132\text{M}\end{array}$

The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation,

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)\\ \text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{NaNO}}_{\text{2}}\right]}{\left[{\text{HNO}}_2\right]}\end{array}$

Substitute the values in the above equation to find the $\text{pH}$ after the addition of $\text{HCl}$.

$\begin{array}{c}\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{NaNO}}_{\text{2}}\right]}{\left[{\text{HNO}}_2\right]}\\ =3.4+\log \frac{0.138}{0.132}\\ =3.4+\log 0.138-\log 0.132\\ =\underset{\_}{3.42}\end{array}$

Conclusion

The $\text{pH}$ at ${\text{25}}^{\text{o}}\text{C}$ of $\text{1}\text{.00}\text{L}$ of a buffer that is $\text{0}\text{.120}\text{M}{\text{HNO}}_{\text{2}}$ and $\text{0}\text{.150}\text{M}{\text{NaNO}}_{\text{2}}$ before and after the addition of $\text{1}\text{.00}\text{mL}$ of a $\text{12}\text{.0}\text{M}\text{HCl}$ is $\underset{\_}{3.42}$.