Question

Solid Zinc Hydroxide, when dissolved in water, can show the following equilibria
$Zn{(OH)_{2\left( s \right)}} \rightleftharpoons Zn{(OH)_{2\left( {aq} \right)}}{\text{ }}{k_1} = {10^{ - 6}}M$
$Zn{(OH)_{2(aq)}} \rightleftharpoons Zn{(OH)_{2\left( {aq} \right)}}{\text{ }}{{\text{k}}_2} = {10^2}M$
$Zn{(OH)^ + }_{(aq)} \rightleftharpoons Z{n^{2 + }}_{(aq)} + O{H_{\left( {aq} \right)}}{\text{ }}{{\text{k}}_3} = {10^{ - 9}}M$
What will be the solubility of zinc hydroxide at pH of 6?
A.${10^{ - 4}}M$
B.${10^{ - 6}}M$
C.$1.2 \times {10^{ - 5}}M$
D.$1.2 \times {10^{ - 6}}M$

Answer 1

Hint: Zinc hydroxide is a chemical compound which slightly dissolves in water. Prolonged exposure to Zinc Hydroxide results in eye irritation and chest pain. It is a weak base and white coloured solid compound. When we heat it gives Zinc Oxide.

Complete step by step answer:
We know that Zinc Hydroxide is a chemical compound which is white in colour and has solid state. It is very slightly soluble in water and is a weak base but prolonged exposure to it causes irritation in eyes, chest pain and dry cough. Zinc hydroxide forms a protective layer in water and when it is heated it gives Zinc Oxide. Zinc does not react with oxygen.
Zinc Hydroxide $Zn{(OH)_2}$ is also found in the form of three rare minerals and is amphoteric in nature that means that it acts as both an acid and a base. It is a weak base and so it is used in baby lotions, cosmetics and ointments and is also used to treat various disorders. It has a pH of 8.7
As we know now that zinc hydroxide is not very much soluble in water and it can be made soluble only by lowering or raising the pH value and by calculating the solubility constant then. Thus many experiments have been held to come to a conclusion that sparingly soluble compounds can have derived equilibrium constants for calculating the solubility.
In the above question we calculate the solubility constant for pH value 6
$\begin{gathered}
  Zn{(OH)_{2\left( s \right)}} \rightleftharpoons Zn{(OH)_{2\left( {aq} \right)}}{\text{ }}{k_1} = {10^{ - 6}}M \\
    \\
\end{gathered} $
$Zn{(OH)_{2(aq)}} \rightleftharpoons Zn{(OH)_{2\left( {aq} \right)}}{\text{ }}{{\text{k}}_2} = {10^2}M$
$Zn{(OH)^ + }_{(aq)} \rightleftharpoons Z{n^{2 + }}_{(aq)} + O{H_{\left( {aq} \right)}}{\text{ }}{{\text{k}}_3} = {10^{ - 9}}M$
$s = \left[ {Zn\left( {O{H_2}} \right)(aq)} \right] + Zn{(OH)^ + } + Z{n^{ + 2}}$
$s = {k_1} + \dfrac{{{k_1}{k_2}}}{{\left[ {O{H^ - }} \right]}} + \dfrac{{{k_1}{k_2}{k_3}}}{{{{\left[ {O{H^ - }} \right]}^2}}}$
$s = {10^{ - 6}} + \dfrac{{{{10}^{ - 4}}}}{{\left[ {O{H^ - }} \right]}} + \dfrac{{{{10}^{ - 13}}}}{{{{\left[ {O{H^ - }} \right]}^2}}}$
$For{\text{ }}pH = 6$
$ \Rightarrow {10^{ - 4}}M$

So, the correct answer is option (A).

Note:
$Zn{(OH)_2}$ is used as an adsorbing agent and for other medical purposes and in medicines. It is also used for commercial production and manufacturing of pesticides. It dissolves in excess aqueous ammonia to form colourless water soluble complex unlike other metals like aluminium and lead.