Chapter 16, Problem 109GQ

Interpretation Introduction

Interpretation:

The decreasing order of the ions concetration in the ionization of oxalic acid has to be determined.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

${\text{K}}_{\text{a}}$ is an acid constant for equilibrium reactions.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

${\text{K}}_{\text{b}}$ is a base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

Answer to Problem 109GQ

The decreasing order of the ions concentration is follows.

${\text{H}}_{\text{2}}{\text{O > H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ > H}}_{\text{3}}{\text{O}}^{\text{+}}\equiv {\text{HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{> C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}{\text{ > OH}}^{\text{-}}$

Explanation of Solution

The oxalic acid is a diprotic acid. It ionizes into two steps.

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{(aq) + H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) K}}_{\text{a1}}\text{=5}{\text{.9×10}}^{\text{-2}}\\ {\text{HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{(aq) + H}}_{\text{2}}\text{O( l)}\rightleftharpoons {\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{(aq) + H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) K}}_{\text{a2}}\text{=6}{\text{.4×10}}^{\text{-5}}\\ \end{array}$

From the ionization steps,

0.10 M solution contains the following ions.

${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4,}}{\text{ H}}_{\text{2}}{\text{O, HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{, C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}{\text{, H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{and OH}}^{\text{-}}$

Let’s calculate the ions concentration by using ICE table:

For the first ionization of oxalic acid:

$\begin{array}{l}{\text{Equilibrium H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\rightleftharpoons {\text{HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{(aq) + H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\\ \text{I 0}\text{.10 -- 0 0}\\ \text{C -x -- +x +x}\\ \text{E (0}\text{.10-x) -- x x}\end{array}$

The equilibrium expression:

${\text{K}}_{\text{a1}}\text{=}\frac{{\text{[HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{]}}$

Substitute the values from ICE table

$\text{5}{\text{.9×10}}^{\text{-2}}\text{=}\frac{\text{(x)(x)}}{\text{(0}\text{.10-x)}}$

Here, ${{\text{[H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{]}}_{\text{initial}}{\text{< 100K}}_{a1}\text{(0}\text{.10 < 5}\text{.9)}$

$\begin{array}{c}{\text{x}}^{\text{2}}\text{= (5}{\text{.9×10}}^{\text{-2}}\text{)(0}\text{.10-x)}\\ {\text{x}}^{\text{2}}\text{= 5}{\text{.9×10}}^{\text{-3}}\text{-5}{\text{.9×10}}^{\text{-2}}\text{-x}\end{array}$

All variants are equal to zero.

Therefore,

${\text{x}}^{\text{2}}\text{+5}{\text{.9×10}}^{\text{-2}}\text{ x-5}{\text{.9×10}}^{\text{-3}}\text{= 0}$

Solve the quadratic equation, we get x value.

$\begin{array}{c}\text{x =}\frac{\text{-5}{\text{.9×10}}^{\text{-2}}\text{±}\sqrt{{\text{(5}{\text{.9×10}}^{\text{-2}}\text{)}}^{\text{2}}\text{-4(1)(-5}{\text{.9×10}}^{\text{-3}}\text{)}}}{\text{2(1)}}\\ \text{x =}\frac{\text{-5}{\text{.9×10}}^{\text{-2}}\text{±}\sqrt{\text{34}{\text{.81×10}}^{\text{-4}}\text{+23}{\text{.6×10}}^{\text{-3}}}}{\text{2}}\\ \text{x = 0}\text{.053 M}\end{array}$

Therefore,

$\begin{array}{c}{\text{[HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}\text{] =0}\text{.053 M}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] =0}\text{.053 M}\\ {\text{[H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{] =(0}\text{.10-0}\text{.053) M}\\ \text{= 0}\text{.047 M}\end{array}$

Let’s calculate the ion concentration from second ionization of oxalic acid.

$\begin{array}{l}{\text{Equilibrium HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}{\text{(aq) + H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\\ \text{I 0}\text{.053 -- 0 0}\text{.053}\\ \text{C -y -- +y +y}\\ \text{E (0}\text{.053-y) -- y (0}\text{.053 + y)}\end{array}$

The equilibrium expression:

${\text{K}}_{\text{a2}}\text{= }\frac{{\text{[C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}\text{]}}$

Substitute the values from the ICE table:

$\text{6}{\text{.4×10}}^{\text{-5}}\text{=}\frac{\text{y(0}\text{.053+y)}}{\text{(0}\text{.053-y)}}$

${\text{K}}_{\text{a2}}$ is very small therefore,

$\text{(0}\text{.053-y) and (0}\text{.053 + y)are equal to 0}\text{.053}$

$\begin{array}{c}\text{6}{\text{.4×10}}^{\text{-5}}\text{ = }\frac{\text{y(0}\text{.053)}}{\text{(0}\text{.053)}}\\ \text{y = 6}{\text{.4×10}}^{\text{-5}}\end{array}$

Therefore,

${\text{[C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\text{] = 6}{\text{.4×10}}^{\text{-5}}\text{ M}$

The amount of hydronium ions are produced in the second step ionization is negligible.

From the above calculation write the each ion concentration:

$\begin{array}{c}{\text{[HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}\text{]= 0}\text{.053 M}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]= 0}\text{.053 M}\\ {\text{[H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{]= 0}\text{.047 M}\\ {\text{[C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\text{] = 6}{\text{.4×10}}^{\text{-5}}\text{ M}\end{array}$

Let’s calculate the hydroxide ion concentration.

$\begin{array}{c}{\text{K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{= 1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]= }\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{0}\text{.053}}\\ {\text{[OH}}^{\text{-}}\text{]= 1}{\text{.9×10}}^{\text{-13}}\text{ M}\end{array}$

The decreasing order of the ions concentration is follows.

${\text{H}}_{\text{2}}{\text{O > H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ > H}}_{\text{3}}{\text{O}}^{\text{+}}\equiv {\text{HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{> C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}{\text{ > OH}}^{\text{-}}$

Conclusion

Therefore, the decreasing order of the ions concentration is follows.

${\text{H}}_{\text{2}}{\text{O > H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ > H}}_{\text{3}}{\text{O}}^{\text{+}}\equiv {\text{HC}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{-}}{\text{> C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}{\text{ > OH}}^{\text{-}}$