Chapter 1, Problem 18P

Interpretation Introduction

(a)

Interpretation:

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ at $\text{pH}$ 7.0 is to be determined.

Concept introduction:

Henderson and Haselbalch equation is used to find the $\text{pH}$. The expression is as given below,

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

Where,

• $\text{p}{K}_{\text{a}}$ is dissociation constant.
• $\left[\text{Salt}\right]$ is the concentration of salt.
• $\left[\text{Acid}\right]$ is the concentration of acid.

Answer to Problem 18P

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ at $\text{pH}$ 7.0 is 1.62.

Explanation of Solution

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ is calculated by the following equation:

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}$ ...... (1)

Rearrange equation (1) as follows:

$\log \frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}=\text{pH}-\text{p}{K}_{\text{a}}$ ...... (2)

The value of $\text{pH}$ is 7.

The value of $\text{p}{K}_{\text{a}}$ is 7.21.

Substitute the values of $\text{pH}$ and $\text{p}{K}_{\text{a}}$ in equation (2),

$\begin{array}{c}\log \frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}=7.21-7.0\\ =0.21\end{array}$

Solve for $\frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}$ ,

$\begin{array}{c}\frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}={10}^{0.21}\\ =1.62\end{array}$

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ is 1.62.

Interpretation Introduction

(b)

Interpretation:

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ at $\text{pH}$ 7.5 is to be determined.

Concept introduction:

Henderson and Haselbalch equation is used to find the $\text{pH}$. The expression is as given below,

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

Where,

• $\text{p}{K}_{\text{a}}$ is dissociation constant.
• $\left[\text{Salt}\right]$ is the concentration of salt.
• $\left[\text{Acid}\right]$ is the concentration of acid.

Answer to Problem 18P

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ at pH 7.5 is 0.51.

Explanation of Solution

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ can be given by the help of following equation,

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}$

Substitute the values of $\text{pH}$ and $\text{p}{K}_{\text{a}}$ in the above equation,

$\begin{array}{c}7.5=7.21+\log \frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}\\ 0.29=\log \frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}\end{array}$

Rearrange the above equation for $\frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}$

$\begin{array}{c}\frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}={10}^{-0.29}\\ =0.51\end{array}$

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ is 0.51.

Interpretation Introduction

(c)

Interpretation:

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ at $\text{pH}$ 8.0 is to be determined.

Concept introduction:

Henderson and Haselbalch equation is used to find the $\text{pH}$. The expression is as given below,

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

Where,

• $\text{p}{K}_{\text{a}}$ is dissociation constant.
• $\left[\text{Salt}\right]$ is the concentration of salt.
• $\left[\text{Acid}\right]$ is the concentration of acid.

Answer to Problem 18P

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ at $\text{pH}$ 8.0 is 0.16.

Explanation of Solution

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ can be given by the help of following equation,

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}$

Substitute the values of $\text{pH}$ and $\text{p}{K}_{\text{a}}$ in the above equation,

$\begin{array}{c}8.0=7.21+\log \frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}\\ 0.79=\log \frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}\end{array}$

Rearrange the above equation for $\frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}$

$\begin{array}{c}\frac{\left[{\text{HPO}}_4{}^{2-}\right]}{\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]}={10}^{-0.79}\\ =0.16\end{array}$

The ratio of concentrations of ${\text{H}}_2{\text{PO}}_4{}^{-}$ and ${\text{HPO}}_4{}^{2-}$ is 0.16.