Chapter 2, Problem 8P

Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.

Polyprotic Adds: Phosphate Species Abundance at Different pHs What are the approximate fractional concentrations of the following phosphate species at pH values of 0,2, 4, 6, 8, 10, and 12?

1. H₃PO₄
2. Η₂ΡΟ₄-
3. HPO₄^2-
4. PO₄^3-

Interpretation Introduction

(a)

To calculate:

The approximate fractional concentrations of H₃PO₄ at pH values of 0,2,4,6,8,10, and 12.

Concept Introduction:

Phosphoric acid has three H+ dissociations.

  $\begin{array}{l}{\text{H}}_3{\text{PO}}_4\rightleftharpoons {\text{H}}^{+}+{\text{H}}_2{\text{PO}}_4^{-}{\text{ pK}}_{a1}=2.15\\ {\text{H}}_2{\text{PO}}_4^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HPO}}_4^{2-}{\text{ pK}}_{a2}=7.20\\ {\text{HPO}}_4^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{PO}}_4^{3-}{\text{ pK}}_{a3}=12.4\end{array}$

The Henderson-Hasselbalch equation is as follows:

  $pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

Explanation of Solution

Rearranging the Henderson-Hasselbalch equation as follows:

  $\begin{array}{l}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}\\ \frac{\left[A^{-}\right]}{\left[HA\right]}=anti\log \left(pH-p{K}_a\right)\end{array}$

At particular pH, $\frac{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}\text{ = x, }\frac{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}\text{ = y, }\frac{\left[{\text{PO}}_{\text{4}}^{\text{3-}}\right]}{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}\text{ = z}$

Fraction of any species at a particular pH is concentration of the species divided by the sum of the concentration of all species.

  $f_{H_{3}P{O}_4}=\frac{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]+\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]+\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]+\left[{\text{PO}}_{\text{4}}^{\text{3-}}\right]}$

We can rewrite the above equation as a function of x, y, and z.

  $\begin{array}{l}{f}_{H_{3}P{O}_4}=\frac{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]+x\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]+xy\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]+xyz\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}\\ f_{H_{3}P{O}_4}=\frac{1}{1+x+xy+xyz}\end{array}$

We can manually calculate x, y, z and fractions of each species, but it is time consuming. Therefore, for efficiency we can calculate them using a spread sheet as follows:

pH	x	y	z
0	0.00707946	6.3096E-08	3.9811E-13
2	0.70794578	6.3096E-06	3.9811E-11
4	70.7945784	0.00063096	3.9811E-09
6	7079.45784	0.06309573	3.9811E-07
8	707945.784	6.30957344	3.9811E-05
10	70794578.4	630.957344	0.00398107
12	7079457844	63095.7344	0.39810717

Thus,

pH	Fraction of H3PO4
0	0.993
2	0.585
4	0.014
6	0.000
8	0.000
10	0.000
12	0.000

Interpretation Introduction

(b)

To calculate:

The approximate fractional concentrations of H₂PO₄- at pH values of 0,2,4,6,8,10, and 12 should be calculated

Introduction:

Phosphoric acid has three H+ dissociations.

  $\begin{array}{l}{\text{H}}_3{\text{PO}}_4\rightleftharpoons {\text{H}}^{+}+{\text{H}}_2{\text{PO}}_4^{-}{\text{ pK}}_{a1}=2.15\\ {\text{H}}_2{\text{PO}}_4^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HPO}}_4^{2-}{\text{ pK}}_{a2}=7.20\\ {\text{HPO}}_4^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{PO}}_4^{3-}{\text{ pK}}_{a3}=12.4\end{array}$

The Henderson-Hasselbalch equation is as follows:

  $pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

Explanation of Solution

Rearranging the Henderson-Hasselbalch equation as follows:

  $\begin{array}{l}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}\\ \frac{\left[A^{-}\right]}{\left[HA\right]}=anti\log \left(pH-p{K}_a\right)\end{array}$

At particular pH, $\frac{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}\text{ = x, }\frac{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}\text{ = y, }\frac{\left[{\text{PO}}_{\text{4}}^{\text{3-}}\right]}{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}\text{ = z}$

Fraction of any species at a particular pH is concentration of the species divided by the sum of the concentration of all species.

  $f_{{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}}=\frac{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]+\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]+\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]+\left[{\text{PO}}_{\text{4}}^{\text{3-}}\right]}$

We can rewrite the above equation as a function of x, y, and z.

  $\begin{array}{l}{f}_{{\text{H}}_2{\text{PO}}_4^{-}}=\frac{\left[{\text{H}}_2{\text{PO}}_4^{-}\right]}{\frac{\left[{\text{H}}_2{\text{PO}}_4^{-}\right]}{x}+\left[{\text{H}}_2{\text{PO}}_4^{-}\right]+y\left[{\text{H}}_2{\text{PO}}_4^{-}\right]+yz\left[{\text{H}}_2{\text{PO}}_4^{-}\right]}\\ f_{{\text{H}}_2{\text{PO}}_4^{-}}=\frac{1}{\frac{1}{x}+1+y+yz}\end{array}$

We can manually calculate x, y, z and fractions of each species, but it is time consuming. Therefore, for efficiency we can calculate them using a spread sheet as follows:

pH	x	y	z
0	0.00707946	6.3096E-08	3.9811E-13
2	0.70794578	6.3096E-06	3.9811E-11
4	70.7945784	0.00063096	3.9811E-09
6	7079.45784	0.06309573	3.9811E-07
8	707945.784	6.30957344	3.9811E-05
10	70794578.4	630.957344	0.00398107
12	7079457844	63095.7344	0.39810717

Thus,

pH	Fraction of H2PO4-
0	0.007
2	0.415
4	0.985
6	0.941
8	0.137
10	0.002
12	0.000

Interpretation Introduction

(c)

To calculate:

The approximate fractional concentrations of HPO₄²- at pH values of 0,2,4,6,8,10, and 12.

Introduction:

Phosphoric acid has three H+ dissociations.

  $\begin{array}{l}{\text{H}}_3{\text{PO}}_4\rightleftharpoons {\text{H}}^{+}+{\text{H}}_2{\text{PO}}_4^{-}{\text{ pK}}_{a1}=2.15\\ {\text{H}}_2{\text{PO}}_4^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HPO}}_4^{2-}{\text{ pK}}_{a2}=7.20\\ {\text{HPO}}_4^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{PO}}_4^{3-}{\text{ pK}}_{a3}=12.4\end{array}$

The Henderson-Hasselbalch equation is as follows:

  $pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

Explanation of Solution

Rearranging the Henderson-Hasselbalch equation as follows:

  $\begin{array}{l}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}\\ \frac{\left[A^{-}\right]}{\left[HA\right]}=anti\log \left(pH-p{K}_a\right)\end{array}$

At particular pH, $\frac{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}\text{ = x, }\frac{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}\text{ = y, }\frac{\left[{\text{PO}}_{\text{4}}^{\text{3-}}\right]}{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}\text{ = z}$

Fraction of any species at a particular pH is concentration of the species divided by the sum of the concentration of all species.

  $f_{{\text{HPO}}_{\text{4}}^{\text{2-}}}=\frac{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]+\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]+\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]+\left[{\text{PO}}_{\text{4}}^{\text{3-}}\right]}$

We can rewrite the above equation as a function of x, y, and z.

  $\begin{array}{l}{f}_{{\text{HPO}}_4^{2-}}=\frac{\left[{\text{HPO}}_4^{2-}\right]}{\frac{\left[{\text{HPO}}_4^{2-}\right]}{xy}+\frac{\left[{\text{HPO}}_4^{2-}\right]}{y}+\left[{\text{HPO}}_4^{2-}\right]+z\left[{\text{HPO}}_4^{2-}\right]}\\ f_{{\text{HPO}}_4^{2-}}=\frac{1}{\frac{1}{xy}+\frac{1}{y}+1+z}\end{array}$

We can manually calculate x, y, z and fractions of each species, but it is time consuming. Therefore, for efficiency we can calculate them using a spread sheet as follows:

pH	x	y	z
0	0.00707946	6.3096E-08	3.9811E-13
2	0.70794578	6.3096E-06	3.9811E-11
4	70.7945784	0.00063096	3.9811E-09
6	7079.45784	0.06309573	3.9811E-07
8	707945.784	6.30957344	3.9811E-05
10	70794578.4	630.957344	0.00398107
12	7079457844	63095.7344	0.39810717

Thus,

pH	Fraction of HPO42-
0	0.000
2	0.000
4	0.001
6	0.059
8	0.863
10	0.994
12	0.715

Interpretation Introduction

(d)

To calculate:

The approximate fractional concentrations of PO₄³- at pH values of 0,2,4,6,8,10, and 12.

Concept Introduction:

Phosphoric acid has three H+ dissociations.

  $\begin{array}{l}{\text{H}}_3{\text{PO}}_4\rightleftharpoons {\text{H}}^{+}+{\text{H}}_2{\text{PO}}_4^{-}{\text{ pK}}_{a1}=2.15\\ {\text{H}}_2{\text{PO}}_4^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HPO}}_4^{2-}{\text{ pK}}_{a2}=7.20\\ {\text{HPO}}_4^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{PO}}_4^{3-}{\text{ pK}}_{a3}=12.4\end{array}$

The Henderson-Hasselbalch equation is as follows:

  $pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

Explanation of Solution

Rearranging the Henderson-Hasselbalch equation as follows:

  $\begin{array}{l}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}\\ \frac{\left[A^{-}\right]}{\left[HA\right]}=anti\log \left(pH-p{K}_a\right)\end{array}$

At particular pH, $\frac{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}\text{ = x, }\frac{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}\text{ = y, }\frac{\left[{\text{PO}}_{\text{4}}^{\text{3-}}\right]}{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}\text{ = z}$

Fraction of any species at a particular pH is concentration of the species divided by the sum of the concentration of all species.

  $f_{{\text{PO}}_{\text{4}}^{\text{3-}}}=\frac{\left[{\text{PO}}_{\text{4}}^{\text{3-}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]+\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]+\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]+\left[{\text{PO}}_{\text{4}}^{\text{3-}}\right]}$

We can rewrite the above equation as a function of x, y, and z.

  $\begin{array}{l}{f}_{{\text{PO}}_4^{3-}}=\frac{\left[{\text{PO}}_4^{3-}\right]}{\frac{\left[{\text{PO}}_4^{3-}\right]}{xyz}+\frac{\left[{\text{PO}}_4^{3-}\right]}{yz}+\frac{\left[{\text{PO}}_4^{3-}\right]}{z}+\left[{\text{PO}}_4^{3-}\right]}\\ f_{{\text{PO}}_4^{3-}}=\frac{1}{\frac{1}{xyz}+\frac{1}{yz}+\frac{1}{z}+1}\end{array}$

We can manually calculate x, y, z and fractions of each species, but it is time consuming. Therefore, for efficiency we can calculate them using a spread sheet as follows:

pH	x	y	z
0	0.00707946	6.3096E-08	3.9811E-13
2	0.70794578	6.3096E-06	3.9811E-11
4	70.7945784	0.00063096	3.9811E-09
6	7079.45784	0.06309573	3.9811E-07
8	707945.784	6.30957344	3.9811E-05
10	70794578.4	630.957344	0.00398107
12	7079457844	63095.7344	0.39810717

Thus,

pH	Fraction of PO43-
0	0.000
2	0.000
4	0.000
6	0.000
8	0.000
10	0.004
12	0.285