Question

Calculate the solubility of solid zinc hydroxide at a pH of 5.
Given:
\[
  Zn{\left( {OH} \right)_2}\left( s \right) \rightleftarrows Zn{\left( {OH} \right)_2}\left( {aq} \right){k_1} = {10^{ - 6}}M...........(1) \\
  Zn{\left( {OH} \right)_2}\left( {aq} \right) \rightleftarrows Zn{\left( {OH} \right)^ + } + O{H^ - }{k_2} = {10^{ - 7}}M...........(2) \\
  Zn{\left( {OH} \right)^ + } \rightleftarrows Z{n^2} + O{H^ - }{k_3} = {10^{ - 4}}M...........(3) \\
  Zn{\left( {OH} \right)_2}\left( {aq} \right) + O{H^ - } \rightleftarrows Zn{\left( {OH} \right)_3}^ - {k_4} = {10^3}{M^{ - 1}}...........(4) \\
  Zn{\left( {OH} \right)_3} + O{H^ - } \rightleftarrows Zn{\left( {OH} \right)_4}^{2 - }{k_5} = 10{M^{ - 1}}...........(5) \\
\]

Answer 1

Hint: For the given question we need to take account of following things: solubility constant ${K_p}$ and $pH$, the relationship involving ${K_p}$ and $pH$ for the given question is:
\[s = {k_1} + \dfrac{{{k_1}{k_2}}}{{[O{H^ - }]}} + \dfrac{{{k_1}{k_2}{k_3}}}{{{{[O{H^ - }]}^2}}} + {k_1}{k_4}\left[ {O{H^ - }} \right] + {k_1}{k_4}{k_5}{\left[ {O{H^ - }} \right]^2}\]
Here \[s\] is the solubility, $k$ is the rate constant and \[O{H^ - }\] represents the hydroxide ions.

Complete step by step answer:
Let’s 1st understand the meaning of solubility, it is defined as the product of concentrations of the ions raised to their power equal to the number of times the ions occurs in the equation which represents the dissociation of electrolyte at a given temperature when the solution is saturated.
Now as per the given question we can write,
$s = [Zn{\left( {OH} \right)_2}\left( {aq} \right)] + Zn{\left( {OH} \right)^ + } + Z{n^{2 + }} + Zn{\left( {OH} \right)_3}^ - + Zn{\left( {OH} \right)_4}^{2 - }$
We know that,
Relationship between ${K_p}$and $pH$ is given by:
\[s = {k_1} + \dfrac{{{k_1}{k_2}}}{{[O{H^ - }]}} + d\frac{{{k_1}{k_2}{k_3}}}{{{{[O{H^ - }]}^2}}} + {k_1}{k_4}\left[ {O{H^ - }} \right] + {k_1}{k_4}{k_5}{\left[ {O{H^ - }} \right]^2}\]
Substituting the value of rate constants as given in question to the above equation we get,
\[s = {10^{ - 6}} + \dfrac{{{{10}^{ - 13}}}}{{[O{H^ - }]}} + \dfrac{{{{10}^{ - 17}}}}{{{{[O{H^ - }]}^2}}} + {10^{ - 3}}[O{H^ - }] + {10^{ - 2}}{[O{H^ - }]^2}\]
Therefore solubility at $pH = 5$, \[[O{H^ - }] = {10^{ - 9}}\] is
$
 s = {10^{ - 6}} + \dfrac{{{{10}^{ - 13}}}}{{[{{10}^{ - 9}}]}} + \dfrac{{{{10}^{ - 17}}}}{{{{[{{10}^{ - 9}}]}^2}}} + {10^{ - 3}}[{10^{ - 9}}] + {10^{ - 2}}{[{10^{ - 9}}]^2} \\
   \Rightarrow s = 10{\text{M}} \\
$
Therefore solubility at $pH = 5$ is \[s = 10{\text{M}}\].

Additional information:
Some of the salts like ${\text{AgI, Pb}}{{\text{I}}_{\text{2}}}$ etc., are considered insoluble, but they have little amount of solubility. These salts are called sparingly soluble electrolyte. A solution of this contains a very small amount of dissolved electrolyte. And it is assumed that this dissolved electrolyte is present in the form of ions, i.e. it is completely dissociated.
Solubility products are not the ionic product under all conditions but only when the solution is saturated. And for precipitation of an electrolyte, it is required that the ionic product must exceed its solubility product.


Note:
For simultaneous solubility, its calculation is divided into 2 cases
Case I: When the 2 electrolyte are almost equally strong
Case II: When the 2 electrolyte are not equally strong
This needs to be taken care of while solving questions.