Question

If only 0.186 g of Ca(OH)2 dissolves in enough water to give 0.230 L of aqueous solution at a given temperature, what is the Ksp value for calcium hydroxide at this temperature?

Answer 1

Answer:

$Ksp=5.20x10^{-6}$

Explanation:

Hello there!

In this case, according to the solubility equilibrium of calcium hydroxide:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

Whereas the equilibrium expression is:

$Ksp=[Ca^{2+}][OH^-]^2$

It is firstly necessary to calculate the molar solubility given the grams and volume of the dissolved solute:

$s=\frac{0.186g/(74.09g/mol)}{0.230L}=0.0109M$

Now, according to the Ksp expression, we plug in s as the solubility to obtain:

$Ksp=(s)(2s)^2\\\\Ksp=(0.109)(2*0.0109)^2\\\\Ksp=5.20x10^{-6}$

Regards!