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Question: Calculate the solubility of solid zinc hydroxide at a pH of 5. Given: \[ Zn{\left( {OH} \rig...

Calculate the solubility of solid zinc hydroxide at a pH of 5.
Given:

Zn(OH)2(s)⇄Zn(OH)2(aq)k1=10−6M...........(1) Zn(OH)2(aq)⇄Zn(OH)++OH−k2=10−7M...........(2) Zn(OH)+⇄Zn2+OH−k3=10−4M...........(3) Zn(OH)2(aq)+OH−⇄Zn(OH)3−k4=103M−1...........(4) Zn(OH)3+OH−⇄Zn(OH)42−k5=10M−1...........(5)  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) \\\
Explanation

Solution

For the given question we need to take account of following things: solubility constant Kp{K_p} and pHpH, the relationship involving Kp{K_p} and pHpH for the given question is:
s=k1+k1k2[OH−]+k1k2k3[OH−]2+k1k4[OH−]+k1k4k5[OH−]2s = {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 ss is the solubility, kk is the rate constant and OH−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(OH)2(aq)]+Zn(OH)++Zn2++Zn(OH)3−+Zn(OH)42−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 Kp{K_p}and pHpH is given by:
s=k1+k1k2[OH−]+dk1k2k3[OH−]2+k1k4[OH−]+k1k4k5[OH−]2s = {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+10−13[OH−]+10−17[OH−]2+10−3[OH−]+10−2[OH−]2s = {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=5pH = 5, [OH−]=10−9[O{H^ - }] = {10^{ - 9}} is
s=10−6+10−13[10−9]+10−17[10−9]2+10−3[10−9]+10−2[10−9]2 ⇒s=10M  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=5pH = 5 is s=10Ms = 10{\text{M}}.

Additional information:
Some of the salts like AgI, PbI2{\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.