Question

The Ksp value is 3.0x10^-16 for Zn(OH)2. Calculate the molar solubility of Zn(OH)2 at room temperature in the solution of 0.0100 M KCl (inert to any reaction with Zn(OH)2) using (1) Activities and (2) molar concentration.  

Answer 1

Before going on to solve the problem, let us know about the factors that do affect the Ksp values. Ksp of a salt is affected by pH of the solution and the formation of complexes.

pH of a solution: A compound with basic anion will be more soluble in a medium whose pH is lower. We know that KCl is a salt of strong acid HCl and a strong base KOH. The solution will therefore be neutral. K ions and Cl ions won't undergo hydrolysis. Thus presence of KCl will not affect the solublity of zinc hydroxide as far as pH is concerned.

Now the next thing to focus on is complex formation. Zn(OH)2 forms a complex in basic medium i.e., a medium with excess of OH- ions.

Zn(OH)2 + 2OH- Zn(OH)₄^2-

This is not the case with KCl solution.(as it is neutral pH~7).

So we can just write the equation for solublity as:

Zn(OH)2 Zn²⁺ + 2OH^-

^{X 2X}

^{X is the molar solublity of Zn(OH)2.}

Ksp = [Zn2+].[OH-]² = X* (2X)^2= 4X^3= 3.0*10^-16

X= 4.2*10^-6 M.

Now considering the activities:

We need to consider the ionic strength of the medium.

I = concentration of KCl as both the ions are univalent

I= 0.01

The activity coefficients of Zn ion is 0.676 and hydroxide ion is 0.900 when ionic strenght of medium is 0.01.

Ksp' = a_Zn2+.a_OH-² = _Zn2+*[Zn2+]*(_OH-.[OH-])²= _Zn2+._OH-²*Ksp

= 0.676*0.900^2*3*10^-16= 1.643*10^-16 M³

a_Zn2+ = 0.676*4.2*10^-6= 2.839*10^-6 M

So the molar solublity in terms of activity is 2.839*10^-6M and in terms of molar concentration is 4.2*10^-6M.

hope this helped you. Have a good day!