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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?

Respuesta :

sebassandin

Answer:

[tex]Ksp=5.20x10^{-6}[/tex]

Explanation:

Hello there!

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

[tex]Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-[/tex]

Whereas the equilibrium expression is:

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

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

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

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

[tex]Ksp=(s)(2s)^2\\\\Ksp=(0.109)(2*0.0109)^2\\\\Ksp=5.20x10^{-6}[/tex]

Regards!

Cricetus

The Ksp value will be "5.20×10⁻⁶".

According to the solution equilibrium,

  • [tex]Ca(OH)_2 \rightleftharpoons Ca^{2+}+2OH^-[/tex]

Now,

The molar solubility will be:

→ [tex]s = \frac{\frac{0.186}{74.09} }{0.230}[/tex]

     [tex]= 0.0109 \ M[/tex]

hence,

The Ksp value will be:

→ [tex]Ksp = (s)(2s)^2[/tex]

By substituting the values,

          [tex]= (0.109)(2\times 0.0109)^2[/tex]

          [tex]= 5.20\times 10^{-6}[/tex]

Thus the above approach is right.

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