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Dmitry [639]
3 years ago
13

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?
Chemistry
1 answer:
Lilit [14]3 years ago
8 0

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!

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