If the molar solubility of CaF_2 at 35 C is 1.24 * 10^-3 mol/L, what is Ksp at this temperature?

1 Answer
anor277
Jun 9, 2016

K_(sp) = [Ca^(2+)][F^-]^2 = ??

Explanation:

K_(sp), "the solubility product" derives from the solubility expression:

CaF_2(s) rightleftharpoonsCa^(2+) + 2F^-

And K_(eq) = ([Ca^(2+)][F^-]^2)/([CaF_2(s)])

However, [CaF_2(s)], the concentration of a solid is a meaningless quantity.

Thus: K_(eq) = K_(sp) = [Ca^(2+)][F^-]^2

Given this expression, we simply fill in the blanks:

K_(sp) = (1.24xx10^-3)(2xx1.24xx10^-3)^2

= 4xx(1.24xx10^-3)^3

= ???

The given K_(sp) is calculated for 35 ""^@C. At lower temperature, would you expect K_(sp) to increase or decrease? Give a reason for your answer.

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