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Question: The solubility product of \( Zn{(OH)_2} \) is \( 10 \) at \( 25^\circ C \) . What would be the conce...

The solubility product of Zn(OH)2Zn{(OH)_2} is 1010 at 25∘C25^\circ C . What would be the concentration Zn+2Z{n^{ + 2}} ion in 0.1M0.1M - NH4OHN{H_4}OH solution which is 50%50\% ionized?
(A). 2×10−132 \times {10^{ - 13}}
(B). 4×10−124 \times {10^{ - 12}}
(C). 4×10−84 \times {10^{ - 8}}
(D). 2×10−112 \times {10^{ - 11}}

Explanation

Solution

Solubility product is a special type of equilibrium that falls under the category of dynamic equilibrium. And mostly found in the solid state compounds. The literal meaning of the term solubility is the ability of a complex ( in case of solubility product is mostly limited to solids only ) to get mixed with one another. The solubility product is also called sparingly soluble salts.

Complete answer:
The solubility product at a given condition of temperature for any salt is defined as the, equal to the multiplied product of the concentrations of the ions in the saturated solution, with each concentration term raised to the power equal to the number of ions produced on dissociation of one mole of the substance. The Ksp{K_{sp}} is called the solubility product or the solubility product constant.
The reaction of Zn(OH)2Zn{(OH)_2} is as follows
Zn(OH)2⇄ Zn++2OH−Zn{(OH)_2} \rightleftarrows \,Z{n^ + } + 2O{H^ - }
S        2SS\,\,\,\,\,\,\,\,2S
By using the general formula for the calculation of Ksp{K_{sp}} = [Ay+]x[Bx−]y{[{A^{y + }}]^x}{[{B^{x - }}]^y}
Now, if it is dissolved in 0.10.1 M NH4OHN{H_4}OH with 50%50\% dissociation then
\eqalign{ & N{H_4}OH \rightleftarrows N{H_4}^ + + O{H^ - } \cr & 0.1M\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr}
Is the condition at the beginning of the reaction. Now taking the condition of equilibrium
\eqalign{ & N{H_4}OH \rightleftarrows N{H_4}^ + + O{H^ - } \cr & 0.5M\,\,\,\,\,\,\,\,\,\,0.05M\,\,\,\,\,\,\,0.05M \cr}
The new concentration of [OH−][O{H^ - }] is 2S+0.05=2.774M2S + 0.05 = 2.774M .
So, Ksp=[Zn]2+[OH−]2{K_{sp}} = {[Zn]^{2 + }}{[O{H^ - }]^2}
Now substituting the given concentrations, 10=[Zn]2+[2.774]210 = {[Zn]^{2 + }}{[2.774]^2}
Thus on rearrangement and dividing the given, we get 4×10−12=[Zn]2+4 \times {10^{ - 12}} = {[Zn]^{2 + }}
Thus option (B) is correct.

Note:
The solubility of any reaction depends upon two factors which are as follows, (i) enthalpy of the salt and (ii) salvation enthalpy of the ions. The calculative formula for several solubility product is as follows, Ksp{K_{sp}} = [Ay+]x[Bx−]y{[{A^{y + }}]^x}{[{B^{x - }}]^y}

The equationThe formula
PbCl2⇄Pb2++2Cl−PbC{l_2} \rightleftarrows P{b^{2 + }} + 2C{l^ - }Ksp=[Pb2+][Cl−]2{K_{sp}} = [P{b^{2 + }}]{[C{l^ - }]^2}
BaSO4⇄Ba2++SO42−BaS{O_4} \rightleftarrows B{a^{2 + }} + S{O_4}^{2 - }Ksp=[Ba2+][SO42−]{K_{sp}} = [B{a^{2 + }}][S{O_4}^{2 - }]
Mg(OH)2⇄Mg2++2OH−Mg{(OH)_2} \rightleftarrows M{g^{2 + }} + 2O{H^ - }Ksp=[Mg2+][OH−]2{K_{sp}} = [M{g^{2 + }}]{[O{H^ - }]^2}