# Simultaneous solubility of CaF2 and SrF2

Solubility of two electrolytes having common ion when they are dissolved in solution is called simultaneous solubility.

How do I calculate the simultaneous solubility of the above mentioned salts given that:

$\mathrm {K_{sp}(\ce CaF_2)= 3.9 \times 10^{-11}}$ and $\mathrm {K_{sp}(\ce SrF_2)= 2.9 \times 10^{-9}}$

My attempt:

Let the solubility of $\ce {CaF_2}$ be $x$ and that of $\ce {SrF_2}$ be $y$.

Two equations are obtained:

1. $3.9\times10^{-11} = x(2(x+y))^2$
2. $2.9 \times 10^{-9} = y(2(x+y))^2$

Dividing the two we get, $x=0.013y$ Substituting this value of $x$ in equation $1$, I got $y= 5.2\times 10^{-3}$ which should be the solubility of $\ce {SrF_2}$. However, answer given is $\pu{9\times 10^{-4}M}$.

How do I solve this problem then?

• The concentration of fluoride ion is not what you think it is. You forgot that little 2 in CaF2. Sep 26, 2017 at 19:14
• @IvanNeretin Isn't concentration of fluoride $(x+y)^2$? Sep 26, 2017 at 19:16
• Absolutely not. That wouldn't even match the dimension. Sep 26, 2017 at 19:18
• Better to solve by letting $\ce{[Ca^{2+}]} = x$ and $\ce{[Sr^{2+}]} = y$
– MaxW
Sep 26, 2017 at 19:19
• I dunno. Guess you're supposed to write an answer yourself. Then again, there is not much value in such an answer, as you were doing almost everything right since the beginning. Sep 27, 2017 at 4:39

Calculate the simultaneous solubility of $$\ce{CaF_2}$$ and $$\ce{SrF_2}$$ given that:

$$\mathrm {K_{sp}(\ce CaF_2)= 3.9 \times 10^{-11}}$$ and $$\mathrm {K_{sp}(\ce SrF_2)= 2.9 \times 10^{-9}}$$

Let $$\ce{[Ca^{2+}]} = x$$ and $$\ce{[Sr^{2+}]} = y$$

So the dissolving $$\ce{CaF_2}$$ contributes $$2x$$ of $$\ce{F^-}$$ and the dissolving $$\ce{SrF_2}$$ contributes $$2y$$ of $$\ce{F^-}$$. So the total $$\ce{[F^-]} = 2x + 2y = 2(x+y)$$

Substituting into the two $$\text{K}_{sp}$$ equations leads to "messy" cubic equations which could be solved numerically since there are two equations with two unknowns.

$$3.9\times10^{-11} = x(2(x+y))^2 = 4x(x+y)^2$$
$$2.9 \times 10^{-9} = y(2(x+y))^2 = 4y(x+y)^2$$

But going back to the original $$\text{K}_{sp}$$ equations we have:

$$\ce{[Ca^{2+}][F^-]^2 = 3.9 \times 10^{-11}}$$
$$\ce{[Sr^{2+}][F^-]^2 = 2.9 \times 10^{-9}}$$

and by dividing the two we get $$\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$$ or $$x = 0.0134y$$. Substituting this into the second equation we get:

$$2.9 \times 10^{-9} = 4y(x+y)^2 = 4y(1.0134y)^2 = 4.108y^3$$ or $$y=8.9\times10^{-4}$$

So:

$$\ce{[Sr^{2+}]} = 8.9\times10^{-4}$$
$$\ce{[Ca^{2+}]} = (0.0134)(8.9\times10^{-4}) = 1.2\times10^{-5}$$
$$\ce{[F^-]} = 2(8.9\times10^{-4} + 1.2\times10^{-5}) = 1.8\times10^{-3}$$

Note that this can also be simplified using significant figures a different way. Since the two $$\mathrm {K_{sp}}$$ values only have two significant figures and $$\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$$, we can assume that for all practical purposes that all the $$\ce{F^-}$$ comes from the dissolution of the $$\ce{SrF_2}$$. So:

$$\ce{4[Sr^{2+}]^3} = 2.9\times10^{-9}$$

Thus

$$\ce{[Sr^{2+}]} = 9.0 \times10^{-4}$$
$$\ce{[F^-]} = \sqrt{\frac{2.9\times10^{-9}}{9.0 \times10^{-4}}} = 1.8\times10^{-3}$$
$$\ce{[Ca^{2+}]} = \dfrac{3.9\times10^{-11}}{(1.8\times10^{-3})^2}=1.2\times10^{-5}$$

• Note that the statement "Let the solubility of $\ce{CaF2}$ be $x$ and that of $\ce{SrF2}$ be $y$" by the OP is wrong because it is not $\ce{CaF2}$ that is $x$ and $\ce{SrF2}$ that is $y$ but rather $\ce{[Ca^{2+}]}$ that is $x$ and $\ce{[Sr^{2+}]}$ that is $y$. This leads to the correct value of $\ce{[F^-]}$ which the OP initially had wrong.
– MaxW
Sep 27, 2017 at 17:05
• Yes, I understood why my statement, "let the solubility..." was wrong. Thanks for pointing that out! Sep 27, 2017 at 18:05