Question

Ksp of Mg(OH)2 is 4.0×10-12.The number of moles of Mg2+ ions in one litre of its saturated solution in 0.1 M NaOH is??

Answer 1

Solubility = molarity

Answer 2

Answer: The number of moles of magnesium ions is $4.0\times 10^{-10}mol$

Explanation:

We are given:

$[OH^-]$ from 0.1 NaOH will be 0.1 M because 1 mole of sodium hydroxide produces 1 mole of sodium ions and 1 mole of hydroxide ions

The equation for the ionization of the magnesium hydroxide is given as:

$Mg(OH)_2\leftrightharpoons Mg^{2+}+2OH^-$

                       s      (2s + 0.1)

Expression for the solubility product of $Mg(OH)_2$ will be:

$K_{sp}=[Mg^{2+}][OH^-]^2\\\\K_{sp}=s\times (2s+0.1)^2=4s^3+0.4s^2+0.01s$

We are given:

$K_{sp}=4.0\times 10^{-12}$

Putting values in above equation, we get:

$4.0\times 10^{-12}=4s^3+0.4s^2+0.01s\\\\s=4.0\times 10^{-10}M$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of magnesium ions = $4.0\times 10^{-10}M$

Volume of solution = 1 L

Putting values in above equation, we get:

$4.0\times 10^{-10}M=\frac{\text{Moles of }Mg^{2+}\text{ ions}}{1L}\\\\\text{Moles of }Mg^{2+}\text{ ions}=4.0\times 10^{-10}mol$

Hence, the number of moles of magnesium ions is $4.0\times 10^{-10}mol$;