Chapter 14, Problem 36QAP

There is a buffer system in blood ${\text{H}}_2{\text{PO}}_4{}^{-}-{\text{HPO}}_4{}^{2-}$ that helps keep the blood pH at 7.40. $\left(K_{\text{a}}{\text{H}}_2{\text{PO}}_4{}^{-}=6.2\times {10}^{-8}\right)$.

(a) Calculate the $\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]/\left[{\text{HPO}}_4{}^{2-}\right]$ ratio at the normal pH of blood.

(b) What percentage of the ${\text{HPO}}_4{}^{2-}$ ions are converted to ${\text{H}}_2{\text{PO}}_4{}^{-}$ when the pH is 7.00?

(c) What percentage of the ${\text{H}}_2{\text{PO}}_4{}^{-}$ions are converted to ${\text{HPO}}_4{}^{2-}$ when the pH is 8.00?

Interpretation Introduction

(a)

Interpretation:

The ratio $\frac{\left[H_{2}P{O}_4{}^{-}\right]}{\left[HP{O}_4{}^{2-}\right]}$ should be determined for the blood system.

Concept introduction:

Henderson-Hasselbalch equation is an equation useful to determine the pH of any solution when any acid’s aqueous solution is related to its acid dissociation constant.

Buffer equation is considered as an equation when ionization of a weak acid in the water occurs resulted into the hydronium ion and the conjugate base of that acid.

Answer to Problem 36QAP

The ratio $\frac{\left[H_{2}P{O}_4{}^{-}\right]}{\left[HP{O}_4{}^{2-}\right]}$ is 0.64 for the blood system.

Explanation of Solution

For $H_{2}P{O}_4^{-}$, we have below-mentioned dissociation equation:

$H_{2}P{O}_4^{-}{}_{\left(aq\right)}\rightleftarrows H^{+}{}_{\left(aq\right)}+HP{O}_4^{-2}{}_{\left(aq\right)}$

For acid-dissociation constant of the given ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$ is as mentioned below:

$\begin{array}{l}{K}_a=\frac{\left[H^{+}\right]\left[H_{2}P{O}_4^{-}\right]}{\left[HP{O}_4^{-2}\right]}\\ \frac{\left[H^{+}\right]}{K_a}=\frac{\left[H_{2}P{O}_4^{-}\right]}{\left[HP{O}_4^{-2}\right]}\mathrm{....}(1)\end{array}$

For $H_{2}P{O}_4^{-}$ solution, we have $K_a=6.2\times {10}^{-8}$ and blood pH as 7.40.

Now, we need to determine the concentration of $\left[{\text{H}}^{\text{+}}\right]$ and we can use pH formula for the same.

$\begin{array}{l}{\text{pH=-log}}_{10}\left[H^{+}\right]\\ \text{7}{\text{.40=-log}}_{10}\left[H^{+}\right]\\ \left[H^{+}\right]={10}^{-7.40}\\ \left[H^{+}\right]=3.98\times {10}^{-8}\end{array}$

In equation (1), we are substituting the value of $\left[H^{+}\right]$ as below.

$\begin{array}{l}\frac{\left[H^{+}\right]}{K_a}=\frac{\left[H_{2}P{O}_4^{-}\right]}{\left[HP{O}_4^{-2}\right]}\\ \frac{\left[H_{2}P{O}_4^{-}\right]}{\left[HP{O}_4^{-2}\right]}=\frac{3.98\times {10}^{-8}}{6.2\times {10}^{-8}}\\ \frac{\left[H_{2}P{O}_4^{-}\right]}{\left[HP{O}_4^{-2}\right]}=0.64\end{array}$

Interpretation Introduction

(b)

Interpretation:

The percentage ${\text{HPO}}_4^{-2}$ should be determined whichis converted into ${\text{HPO}}_4^{-2}$ when the pH is 6.80.

Concept introduction:

Henderson-Hasselbalch equation is an equation useful to determine the pH of any solution when any acid’s aqueous solution is related to its acid dissociation constant.

Buffer equation is considered as an equation when ionization of a weak acid in the water occurs resulted in the hydronium ion and the conjugate base of that acid.

Answer to Problem 36QAP

$HP{O}_4^{-2}$ ions as 54% that are converted into $H_{2}P{O}_4^{-}$ when the pH is 6.80.

Explanation of Solution

The pH of the buffer system as 6.80and the ratio of the concentration of $HP{O}_4^{-2}$ and $H_{2}P{O}_4^{-}$ as 0.64.

Now, let us take the concentration of $H_{2}P{O}_4^{-}$ as 1M so that we can determine the concentration of $HP{O}_4^{-2}$ as follow.

$\begin{array}{l}\frac{\left[H_{2}P{O}_4{}^{-}\right]}{\left[HP{O}_4{}^{2-}\right]}=0.64\\ \left[HP{O}_4{}^{2-}\right]=\frac{1}{0.64}\\ \left[HP{O}_4{}^{2-}\right]=1.56M\end{array}$

We have the dissociation equation of $H_{2}P{O}_4^{-}$ as follow:

$\begin{array}{l}{H}_{2}P{O}_4^{-}{}_{\left(aq\right)}\rightleftarrows H^{+}{}_{\left(aq\right)}+HP{O}_4^{-2}{}_{\left(aq\right)}\\ \text{I: 0}\text{.100 0}\text{.156}\\ \text{C: +x -x}\\ \text{E: 0}\text{.100+x 0}\text{.100-x}\end{array}$

As per the Henderson-Hasselbalch equation,

$\begin{array}{l}{\text{pH= pK}}_a\text{+ log}\frac{\left[base\right]}{\left[acid\right]}\\ {\text{pH= pK}}_a\text{+ log}\frac{\left[HP{O}_4^{-2}\right]}{\left[H_{2}P{O}_4^{-}\right]}\\ \text{6}{\text{.80= -log}}_{10}\left(6.2\times {10}^{-8}\right){\text{+log}}_{10}\frac{0.156-x}{0.100+x}\\ \text{6}\text{.80= 7}{\text{.21+log}}_{10}\frac{0.156-x}{0.100+x}\\ x=0.0842M\end{array}$

Now, the percentage of $HP{O}_4^{-2}$ which is converted to $H_{2}P{O}_4^{-}$ is,

$\begin{array}{l}\frac{0.0842}{0.156}\times 100\\ =54\%\end{array}$

Interpretation Introduction

(c)

Interpretation:

The percentage $H_{2}P{O}_4^{-}$ should be determined which is converted into $HP{O}_4^{-2}$ when the pH is 7.80.

Concept introduction:

Henderson-Hasselbalch equation is an equation useful to determine the pH of any solution when any acid’s aqueous solution is related to its acid dissociation constant.

Buffer equation is considered as an equation when ionization of a weak acid in the water occurs resulted in the hydronium ion and the conjugated base of that acid.

Answer to Problem 36QAP

$H_{2}P{O}_4^{-}$ as 47% are converted into $HP{O}_4^{-2}$ when the pH is 7.80.

Explanation of Solution

We have the pH of the buffer system as 7.80 and the ratio of the concentration of $HP{O}_4^{-2}$ and $H_{2}P{O}_4^{-}$ as 0.64.

Now, let us take the concentration of $H_{2}P{O}_4^{-}$ as 1M so that we can determine the concentration of $HP{O}_4^{-2}$ as follow.

$\begin{array}{l}\frac{\left[H_{2}P{O}_4{}^{-}\right]}{\left[HP{O}_4{}^{2-}\right]}=0.64\\ \left[HP{O}_4{}^{2-}\right]=\frac{1}{0.64}\\ \left[HP{O}_4{}^{2-}\right]=1.56M\end{array}$

We have the dissociation equation of $H_{2}P{O}_4^{-}$ as follow:

$\begin{array}{l}{H}_{2}P{O}_4^{-}{}_{\left(aq\right)}\rightleftarrows H^{+}{}_{\left(aq\right)}+HP{O}_4^{-2}{}_{\left(aq\right)}\\ \text{I: 0}\text{.100 0}\text{.156}\\ \text{C: -y +y}\\ \text{E: 0}\text{.100-y 0}\text{.100+y}\end{array}$

As per the Henderson-Hasselbalch equation,

$\begin{array}{l}{\text{pH= pK}}_a\text{+ log}\frac{\left[base\right]}{\left[acid\right]}\\ {\text{pH= pK}}_a\text{+ log}\frac{\left[HP{O}_4^{-2}\right]}{\left[H_{2}P{O}_4^{-}\right]}\\ \text{7}{\text{.80= -log}}_{10}\left(6.2\times {10}^{-8}\right){\text{+log}}_{10}\frac{0.156+y}{0.100-y}\\ \text{7}\text{.80= 7}{\text{.21+log}}_{10}\frac{0.156+y}{0.100-y}\\ \text{y= 0}\text{.047 M}\end{array}$

Now, the percentage of $H_{2}P{O}_4^{-}$ which is converted to $HP{O}_4^{-2}$ is,

$\begin{array}{l}\frac{0.047}{1.00}\times 100\\ =47\%\end{array}$