Question

At what pH will \[Cu{{(OH)}_{2}}\] start to precipitate from a solution with \[[C{{u}^{2+}}]\] = 0.0015M?
(\[{{K}_{sp}}\] for \[Cu{{(OH)}_{2}}\] = 1.5 x \[{{10}^{-19}}\])

Answer 1

Hint: Write the solubility product reaction for \[Cu{{(OH)}_{2}}\]. Now write the reaction constant at equilibrium. The solubility constant is given to us as well as the concentration of \[[C{{u}^{2+}}]\]. Substitute both these values in the reaction constant at equilibrium to obtain the concentration of hydroxide ions. With this you can determine the pH of the solution at which \[Cu{{(OH)}_{2}}\] starts precipitating.

Complete step by step solution:
Solubility is defined as the property of a substance (solute) to get dissolved in a solvent in order to form a solution.
The solubility product constant is the equilibrium constant for the dissolution of a solid solute in a solvent to form a solution. It is denoted by the symbol ${{K}_{sp}}$. We can form an idea about the dissolution of an ionic compound by the value of its solubility product.
The concentration of \[[C{{u}^{2+}}]\] = 0.0015 M
Reaction:
$\text{Cu(OH}{{\text{)}}_{\text{2}}}\text{ }\to \text{ C}{{\text{u}}^{\text{2+}}}\text{ + 2O}{{\text{H}}^{-}}$
${{\text{K}}_{\text{sp}}}\text{= }\!\![\!\!\text{ C}{{\text{u}}^{\text{2+}}}\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}$
$\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}\text{ }\!\!]\!\!\text{ = }\sqrt{\dfrac{1.5 \times \text{ 1}{{\text{0}}^{-19}}}{\text{0}\text{.0015}}}$
$[\text{O}{{\text{H}}^{-}}]\text{ = 1} \times 1{{\text{0}}^{-8}}$
$pOH =$ $-\log ([O{{H}^{-}}])$
$pOH =$ $-\log ([1 \times \text{ 1}{{\text{0}}^{-8}}])$ = 8.
$pH =$ $14 - pOH$
So, the value of pH is equal to 14 - 8 = 6.

Therefore, the pH at which \[Cu{{(OH)}_{2}}\] will start to precipitate from a solution with \[[C{{u}^{2+}}]\] = 0.0015M is 6. The correct answer is option (C).

Note: The common ion effect describes the effect of adding a common ion on the ​equilibrium of the new solution. The common ion effect generally decreases ​solubility of a solute. The equilibrium shifts to the left to relieve off the excess product formed.
When a strong electrolyte is added to a solution of weak electrolyte, the solubility of weak electrolyte decreases leading to the formation of precipitate at the bottom of the testing apparatus.