Question

What will happen if the $pH$ of the solution of 0.001M $Mg{(N{O_3})_2}$ is adjusted to $pH\,=\,9$?
[Given that $Ksp$ of $Mg{(OH)_2}$ $ = 8.9 \times {10^{ - 12}}$]
A. ppt will take place
B. ppt will not take place
C. solution will be saturated
D. none of the above

Answer 1

Hint: This question is based on the concept of solubility product and solubility calculation. We can compare Solubility products ($Ksp$) with ionic products to know about the precipitation.

Complete solution:
Let us discuss the solubility product and solubility calculation before going onto the question.
Solubility is generally used for sparingly soluble salts, these are some type solutions where we deal with solubilities
1. Simple solution in ${H_2}O$
2. Effect of common ions on solubility
3. Simultaneous solubility
4. \[Precipitat{e_1} + electrolyt{e_1} \to precipitat{e_2} + electrolyt{e_2}\]
5. Condition for precipitation
6. Solubility in buffer solution
7. Solubility due to complex formation
Solubility Product ($Ksp$) is a type of equilibrium constant, so it will be dependent only on temperature for a particular salt.
For example, Let the salt is \[{A_x}{B_y}\], in solution in water, let the solubility in \[{H_2}O\] =’s’ M, then
\[{A_x}{B_y} \rightleftharpoons x{A^{y + }} + y{B^{ - x}}\] \[{K_{sp}} = {(xs)^x}{(ys)^y} = {x^x}.{y^y}.{(S)^{x + y}}\]

Condition of precipitation: For precipitation ionic product ($Ksp$) should be greater than solubility product $Ksp$.

Mathematical Solution:
Given: $pH = 9$, $Ksp$ of $Mg{(OH)_2}$ $ = 8.9 \times {10^{ - 12}}$
Now we have $pH = 9$, $pOH = 5$
Since $pOH = - {\log _{10}}(O{H^ - })$
$(O{H^ - }) = 1 \times {10^{ - 5}}$
$Mg{(OH)_2} \rightleftharpoons M{g^{2 + }} + 2 {O{H^ - }} $
Ionic product of $Mg{(OH)_2}$$ = \left[ {M{g^{2 + }}} \right] \times {\left[ {O{H^ - }} \right]^2}$
$ = 0.001 \times {({10^{ - 5}})^2}$
$ = 1 \times {10^{ - 13}}{M^3}$
On comparing ionic products with $Ksp$ we come to know that ionic product is smaller than $Ksp$.
So ppt will not take place.

Correct option: B

Note:
Remember the solubility product is a kind of equilibrium constant and its value only depends on temperature. $Ksp$Usually increases with an increase in temperature due to increased solubility.