If the molar solubility of $C a F_{2}$ $CaF_2$ at 35 C is $1.24 \cdot 10^{- 3}$ $1.24 * 10^-3$ mol/L, what is Ksp at this temperature?

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If the molar solubility of $C a F_{2}$ $CaF_2$ at 35 C is $1.24 \cdot 10^{- 3}$ $1.24 * 10^-3$ mol/L, what is Ksp at this temperature?

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Answer 1 · 2016-06-09T17:52:42

$K_{s p}$ $K_(sp)$ $=$ $=$ $[C a^{2 +}] {[F^{-}]}^{2}$ $[Ca^(2+)][F^-]^2$ $=$ $=$ $? ?$ $??$

Explanation:

$K_{s p}$ $K_(sp)$ , $the solubility product$ $"the solubility product"$ derives from the solubility expression:

$C a F_{2} (s) ⇌ C a^{2 +} + 2 F^{-}$ $CaF_2(s) rightleftharpoonsCa^(2+) + 2F^-$

And $K_{e q}$ $K_(eq)$ $=$ $=$ $\frac{[C a^{2 +}] {[F^{-}]}^{2}}{[C a F_{2} (s)]}$ $([Ca^(2+)][F^-]^2)/([CaF_2(s)])$

However, $[C a F_{2} (s)]$ $[CaF_2(s)]$ , the concentration of a solid is a meaningless quantity.

Thus: $K_{e q}$ $K_(eq)$ $=$ $=$ $K_{s p}$ $K_(sp)$ $=$ $=$ $[C a^{2 +}] {[F^{-}]}^{2}$ $[Ca^(2+)][F^-]^2$

Given this expression, we simply fill in the blanks:

$K_{s p}$ $K_(sp)$ $=$ $=$ $(1.24 \times 10^{- 3}) {(2 \times 1.24 \times 10^{- 3})}^{2}$ $(1.24xx10^-3)(2xx1.24xx10^-3)^2$

$=$ $=$ $4 \times {(1.24 \times 10^{- 3})}^{3}$ $4xx(1.24xx10^-3)^3$

$=$ $=$ $? ? ?$ $???$

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

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If the molar solubility of $C a F_{2}$ $CaF_2$ at 35 C is $1.24 \cdot 10^{- 3}$ $1.24 * 10^-3$ mol/L, what is Ksp at this temperature?

1 Answer

Explanation:

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If the molar solubility of $C a F_{2}$ $CaF_2$ at 35 C is $1.24 \cdot 10^{- 3}$ $1.24 * 10^-3$ mol/L, what is Ksp at this temperature?