3 years ago

How do I answer this???

Chemistry

1 answer:

Mumz [18]3 years ago

8 0

Will you have to show me what you're talking

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3 years ago

Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.

madreJ [45]

Answer:

$Molar \ solubility=3.12x10^{-5}M$

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

$CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-$

The equilibrium expression is:

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

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent $x$ is computed as follows:

$3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M$

Thus, the molar solubility equals the reaction extent $x$ , therefore:

$Molar \ solubility=3.12x10^{-5}M$

Regards.

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3 years ago