Chapter 7, Problem 93E

Interpretation Introduction

Interpretation: The concentration ${\text{[H}}^{\text{+}}]$ , ${\text{[OH}}^{\text{-}}]$ , ${\text{[H}}_{\text{3}}{\text{AsO}}_4]$ , ${\text{[H}}_2{\text{AsO}}_{{}_4}^{-}]$ , ${\text{[H}}_2{\text{AsO}}_{{}_4}^{-}]$ and ${\text{[AsO}}_{{}_4}^{3-}]$ in a 0.20 M as arsenic acid solution needs to be determined.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A strong acid ionized completely to form respective ions whereas a weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The number of hydrogen ions given by an acid can be different from other acid. On the basis of number of hydrogen ion given by an acid, they can be classified as monoprotic, diprotic and triprotic acids.

Answer to Problem 93E

• $\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\text{0}\text{.033M}\\ {\text{[H}}_{\text{2}}{\text{AsO}}_{\text{4}}{}^{\text{-}}\text{]=0}\text{.033M}\end{array}$
• ${\text{[H}}_{\text{3}}{\text{AsO}}_4\text{]= 0}\text{.17M}$
• ${\text{[ HAsO}}_{\text{4}}{}^{\text{2-}}\text{]=1}{\text{.7×10}}^{\text{-7}}\text{M}$
• ${\text{[ AsO}}_{\text{4}}{}^{\text{3-}}\text{]=}2.7{\text{×10}}^{\text{-17}}$
• ${\text{[OH}}^{\text{-}}\text{]}=3.2\times {\text{10}}^{\text{-13}}$

Explanation of Solution

Given information:

Concentration of acid = 0.20 M

  $\begin{array}{l}{\text{K}}_{\text{a1}}\text{=5}\text{.5}\times {\text{10}}^{\text{-3}}\\ {\text{K}}_{\text{a2}}\text{=1}\text{.7}\times {\text{10}}^{\text{-7}}\\ {\text{K}}_{\text{a3}}\text{=5}\text{.1}\times {\text{10}}^{\text{-12}}\end{array}$

Arsenic acid is a triprotic acid. It can give three hydrogen ions therefore three acids can formed as given below:

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{AsO}}_4{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{AsO}}_4{{}^{\text{-}}}_{\text{(aq)}}\\ {\text{H}}_{\text{2}}{\text{AsO}}_4{{}^{\text{-}}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ HAsO}}_4{{}^{\text{2-}}}_{\text{(aq)}}\\ {\text{HAsO}}_4{{}^{\text{2-}}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ AsO}}_4{{}^{\text{3-}}}_{\text{(aq)}}\end{array}$

The ${\text{K}}_a$ for all the three steps can be shown as:

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{AsO}}_4{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{AsO}}_4{{}^{\text{-}}}_{\text{(aq)}}\\ {\text{K}}_{\text{a1}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][H}}_{\text{2}}{\text{AsO}}_4{}^{\text{-}}\text{]}}{{\text{[H}}_{\text{3}}{\text{AsO}}_4\text{]}}\\ {\text{H}}_{\text{2}}{\text{AsO}}_4{{}^{\text{-}}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ HAsO}}_4{{}^{\text{2-}}}_{\text{(aq)}}\\ {\text{K}}_{\text{a2}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][ HAsO}}_4{}^{\text{2-}}\text{]}}{{\text{[H}}_{\text{2}}{\text{AsO}}_4{}^{\text{-}}\text{]}}\\ {\text{HAsO}}_4{{}^{\text{2-}}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ AsO}}_4{{}^{\text{3-}}}_{\text{(aq)}}\\ {\text{K}}_{\text{a3}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][AsO}}_4{}^{\text{3-}}\text{]}}{{\text{[HAsO}}_4{}^{\text{2-}}\text{]}}\end{array}$

For step-1:

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{AsO}}_4{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{AsO}}_4{{}^{\text{-}}}_{\text{(aq)}}\\ {\text{K}}_{\text{a1}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][H}}_{\text{2}}{\text{AsO}}_4{}^{\text{-}}\text{]}}{{\text{[H}}_{\text{3}}{\text{AsO}}_4\text{]}}\end{array}$

Make ICE table:

Concentration	${\text{H}}_{\text{3}}{\text{AsO}}_4$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{H}}_{\text{2}}{\text{AsO}}_4{}^{\text{-}}$
Initial	0.20	0	0
Change	-x	x	x
Equilibrium	0.20−x	x	x

Calculate ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ :

  $\begin{array}{l}{\text{K}}_{\text{a1}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][H}}_{\text{2}}{\text{AsO}}_{\text{4}}{}^{\text{-}}\text{]}}{{\text{[H}}_{\text{3}}{\text{AsO}}_{\text{4}}\text{]}}\\ \text{5}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{\text{[x] [x]}}{\text{[0}\text{.20-x]}}\text{(0}\text{.20>>x)}\\ \text{5}{\text{.5×10}}^{\text{-3}}\text{ = }\frac{\text{[x] [x]}}{\text{[0}\text{.20]}}\\ {\text{[x]}}^{\text{2}}\text{ =5}{\text{.5×10}}^{\text{-3}}\text{ ×0}\text{.20}\\ \text{[x]=0}{\text{.033=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]=[H}}_{\text{2}}{\text{AsO}}_{\text{4}}{}^{\text{-}}\text{]}\end{array}$

  ${\text{[H}}_{\text{3}}{\text{AsO}}_4\text{]=0}\text{.20 -0}\text{.033 = 0}\text{.17M}$

Calculate [ ${\text{OH}}^{\text{-}}$ ]:

  $\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{] = 1}\text{.0 }\times {\text{10}}^{\text{-14}}\\ \text{[0}{\text{.033]×[OH}}^{\text{-}}\text{] = 1}\text{.0 }\times {\text{10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]}=\frac{\text{1}\text{.0 }\times {\text{10}}^{\text{-14}}}{\text{[0}\text{.033]}}\\ {\text{[OH}}^{\text{-}}\text{]}=3.2\times {\text{10}}^{\text{-13}}\end{array}$

Use same ${\text{[H}}_{\text{2}}{\text{AsO}}_{\text{4}}{}^{\text{-}}\text{]=0}\text{.033M}$ to calculate ${\text{HAsO}}_4{}^{\text{2-}}$:

Concentration	${\text{H}}_{\text{2}}{\text{AsO}}_4{}^{\text{-}}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{HAsO}}_4{}^{\text{2-}}$
Initial	0.033	0.033	0
Change	-x	0.033	x
Equilibrium	0.033−x	0.033	x

  $\begin{array}{l}{\text{K}}_{\text{a2}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][ HAsO}}_{\text{4}}{}^{\text{2-}}\text{]}}{{\text{[H}}_{\text{2}}{\text{AsO}}_{\text{4}}{}^{\text{-}}\text{]}}\\ \text{1}{\text{.7×10}}^{\text{-7}}\text{=}\frac{\text{[0}{\text{.033][ HAsO}}_{\text{4}}{}^{\text{2-}}\text{]}}{\text{[0}\text{.033-x]}}\text{(0}\text{.033>>x)}\\ {\text{[ HAsO}}_{\text{4}}{}^{\text{2-}}\text{]=1}{\text{.7×10}}^{\text{-7}}\text{M}\end{array}$

Use same ${\text{HAsO}}_4{}^{\text{2-}}$ to calculate ${\text{AsO}}_4{}^{\text{3-}}$:

Concentration	${\text{H}}_{\text{2}}{\text{AsO}}_4{}^{\text{-}}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{AsO}}_4{}^{\text{3-}}$
Initial	$\text{1}{\text{.7×10}}^{\text{-7}}$	0.033	0
Change	-x	0.033	x
Equilibrium	$\text{1}{\text{.7×10}}^{\text{-7}}$ –x	0.033	x

  $\begin{array}{l}{\text{K}}_{\text{a3}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][ AsO}}_{\text{4}}{}^{\text{3-}}\text{]}}{{\text{[HAsO}}_{\text{4}}{}^{\text{2-}}\text{]}}\\ \text{5}{\text{.1×10}}^{\text{-12}}\text{=}\frac{\text{[0}{\text{.033][ AsO}}_{\text{4}}{}^{\text{3-}}\text{]}}{\text{[1}{\text{.7×10}}^{\text{-7}}\text{M-x]}}\text{(1}{\text{.7×10}}^{\text{-7}}\text{M>>x)}\\ {\text{[ AsO}}_{\text{4}}{}^{\text{3-}}\text{]=}\frac{\text{5}{\text{.1×10}}^{\text{-12}}\times \text{1}{\text{.7×10}}^{\text{-7}}}{\text{0}\text{.033}}\\ {\text{[ AsO}}_{\text{4}}{}^{\text{3-}}\text{]=}2.7{\text{×10}}^{\text{-17}}\end{array}$

Conclusion

Thus, concentration of all the species is:

• $\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\text{0}\text{.033M}\\ {\text{[H}}_{\text{2}}{\text{AsO}}_{\text{4}}{}^{\text{-}}\text{]=0}\text{.033M}\end{array}$
• ${\text{[H}}_{\text{3}}{\text{AsO}}_4\text{]= 0}\text{.17M}$
• ${\text{[ HAsO}}_{\text{4}}{}^{\text{2-}}\text{]=1}{\text{.7×10}}^{\text{-7}}\text{M}$
• ${\text{[ AsO}}_{\text{4}}{}^{\text{3-}}\text{]=}2.7{\text{×10}}^{\text{-17}}$
• ${\text{[OH}}^{\text{-}}\text{]}=3.2\times {\text{10}}^{\text{-13}}$