Chapter 18, Problem 18.81P

Interpretation Introduction

Interpretation:

The ${\text{[H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}{\text{], [HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}{\text{], [C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{2-}{\text{], [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{], pH, [OH}}^{\text{-}}\text{], and pOH}$ have to be calculated in $\text{0}\text{.20 M}$ solution of the diprotic malonic acid.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general acid HA,

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

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  $K_a=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$                                                                                                   $\left(1\right)$

An equilibrium constant $\left(K\right)$ with subscript $a$ indicate that it is an equilibrium constant of an acid in water.

Explanation of Solution

Given,

Solution of $\text{0}\text{.2}\text{M}{\text{malonic acid (H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}$.

Malonic acid is diprotic acid therefore it can donate the two protons to the water.

The balance equation is given below,

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\text{ (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\text{(aq)}\\ \\ The value{\text{ K}}_{a1}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{a1}\text{ =}\frac{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\right]}\\ \text{given,}\\ {\text{K}}_{a1}\text{ =}\frac{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\right]}=\text{ 1}\text{.4}\times {10}^{-3}\end{array}$

The formed ${\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}$ ion can donate another proton, therefore the balance equation is given below,

$\begin{array}{l}{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\text{ (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{2-}\text{(aq)}\\ \\ The value{\text{ K}}_{a2}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{a2}\text{ =}\frac{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{2-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]}\\ \text{given,}\\ {\text{K}}_{a2}\text{ =}\frac{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{2-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]}=\text{ 2}\text{.0}\times {10}^{-6}\end{array}$

From the dissociation, ${\text{K}}_{a1}\rangle \rangle {\text{K}}_{a2}$, therefore the first dissociation forms complete ${\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}$ ion.

 $\text{0}\text{.2}\text{M}{\text{malonic acid (H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}$ Solution.

Therefore,

ICE table:

  ${\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\text{ (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\text{(aq)}$

Initial concentration	0.2 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.2-x		x	x

The initial concentration is $\text{0}\text{.2}\text{M}{\text{malonic acid (H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}$.

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\text{ (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\text{(aq)}\\ \\ The value{\text{ K}}_{a1}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{a1}\text{ =}\frac{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\right]}\\ \text{given,}\\ {\text{K}}_{a1}\text{ =}\frac{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\right]}=\text{ 1}\text{.4}\times {10}^{-3}\end{array}$

Let consider,

  $\begin{array}{l}{\text{K}}_{a1}\text{ =}\frac{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\right]}=\text{ 1}\text{.4}\times {10}^{-3}\\ \frac{x^2}{(0.2-x)}=\text{ 1}\text{.4}\times {10}^{-3}\\ x^2=2.8\times {10}^{-4}\\ x=0.0167\end{array}$

Percent dissociation can be calculated by using following formula,

  $\begin{array}{c}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \\ \text{Percent dissociated = }\frac{0.0167}{0.2}\text{×100}\\ \\ \text{Percent dissociated = 8}\text{.3}\%\end{array}$

Percentage of error is high and it is not valid, the dissociation of ${\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}$ is not negligible in comparison to the initial concentration. Therefore, the equilibrium expression can be solved using the quadratic formula.

The quadratic formula is given below,

  $\begin{array}{l}{x}^2+1.4\times {10}^{-3}x-2.8\times {10}^{-4}=0\\ a=1,b=1.4\times {10}^{-3},c=2.8\times {10}^{-4}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(1.4\times {10}^{-3})\pm \sqrt{{(1.4\times {10}^{-3})}^2-4(1)(2.8\times {10}^{-4})}}{2(1)}\\ x=1.6\times {10}^{-2}\text{M}\text{= }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right],\\ \left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]=\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.6\times {10}^{-2}\text{M}\end{array}$

Hence, $\left[H{C}_3{H}_2{O}_4^{-}\right]=\left[H_3{O}^{+}\right]=1.6\times 10^{-2}M$

Therefore,

  $\begin{array}{c}\text{pH}\text{=}\text{-}\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}\text{-}\text{log}\left(1.6\times {10}^{-2}\text{M}\right)\\ \\ \text{pH}\text{=}1.81\end{array}$

 $\text{pH}$ of $\text{0}{\text{.2 M H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}$ is $1.81.$.

The $\left[{\text{OH}}^{\text{-}}\right]$ value is calculating by using following formula,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{ = }\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{ = }\frac{{\text{1×10}}^{\text{-14}}}{1.6\times {10}^{-2}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{ = 6}{\text{.25×10}}^{\text{-13}}\end{array}$

Hence, $\left[{\text{OH}}^{\text{-}}\right]$ is $6.25\times 10^{-13}.$

Therefore,

  $\begin{array}{c}\text{pOH}\text{=}\text{-}\text{log}\left[{\text{OH}}^{-}\right]\\ \\ \text{pOH}\text{=}\text{-}\text{log}\left(\text{6}{\text{.25×10}}^{\text{-13}}\right)\\ \\ \text{pOH}\text{=}12.20\end{array}$

pOH of the solution is $12.20.$

The Malonic acid concentration at equilibrium is ${\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\right]}_{\text{init}}\text{ – }{\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\right]}_{\text{dissoc}}$

The Malonic acid concentration at equilibrium is $\text{0}\text{.2 – 1}\text{.6}\times {10}^{-2}\text{M}$

The Malonic acid concentration at equilibrium is $0.184M.$

Therefore, the ${\text{K}}_{a2}$ expression is given below for calculating ${\text{[C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{2-}\text{]}$, and $\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]\text{ and }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ come mostly from the first dissociation. This new calculation will have a new x value,

Therefore,

ICE table:

  ${\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\text{ (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{2-}\text{(aq)}$

Initial concentration	$1.6\times {10}^{-2}\text{M}$	-	$1.6\times {10}^{-2}\text{M}$	0
Change	-x		+ x	+ x
At equilibrium	$1.6\times {10}^{-2}\text{M}$ -x		$1.6\times {10}^{-2}\text{M}$+x	x

  $\begin{array}{l}The value{\text{ K}}_{a2}iscalculatingbyu\sin gfollowingformula, \\ \\ {\text{K}}_{a2}\text{ =}\frac{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{2-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\right]}=\text{ 2}\text{.0}\times {10}^{-6}\\ \\ {\text{K}}_{a2}\text{=}\frac{\left(1.6\times {10}^{-2}+x\right)\times x}{\left(1.6\times {10}^{-2}-x\right)}=\text{2}\text{.0}\times {10}^{-6}\\ \\ {\text{K}}_{a2}\text{=}\frac{\left(1.6\times {10}^{-2}\right)\times x}{\left(1.6\times {10}^{-2}-x\right)}=2.\text{0}\times {10}^{-5}\\ \\ \left[{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{2-}\right]\text{= }2.\text{0}\times {10}^{-5}\text{M}\end{array}$

X value is very small, hence not necessary to calculate the ${\text{HC}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{-}\text{ and}{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}$.

Hence, $\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{2}}{\text{O}}_4^{2-}\right]$ is $2.0\times 10^{-5}M.$