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Here was my attempt #$1$ at a solution.

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

Using the information at pH=7, I estimated that $K_{sp}\approx3.7\cdot10^{-11}$, which means that barely any HF formed.

I assumed that only $HF$ forms and not $CaOH$ because it's insoluble.

I got $\displaystyle \frac{[H^+][F^-]}{HF}=10^{-3.17}$

We have $ \displaystyle \sqrt{\frac{3.7\cdot10^{-11}}{[Ca^{2+}]}}$ fluoride ions.

We rewrite $K_a=10^{-3.17}=\displaystyle\frac{10^{-3}[F^-]}{2[Ca^{2+}]-F^-}$ and plug in the concentration of fluoride ions in terms of calcium ions.

This yields $[Ca^{2+}]=3.7*10^{-4}$, which gives a multiplier of 1.76.

The correct answer is 1.83, which I guessed based on my answer. How do I arrive at that?

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The $K_{sp}$ of $\ce{CaF2}$ which you found out as $3.7 \times 10^{-11}$ is indeed correct, and after that essentially you have to solve two equations.

Now, suppose at $pH =3$, (i.e. $\ce{[H+] = 10^{-3}}$), $\ce{CaF2}$ dissociates as $\ce{Ca^2+}$ and $\ce{F-}$ and their concentration in the solution are as follows, $$\ce{CaF2 <=> Ca^2+ + 2F-}$$ $\ce{[Ca^2+ ] = S }$ mol/L. and $\ce{[F-] = 2S -}$ $x$ mol/L. ,as, this $x$ mol/L. of $\ce{F-}$ is in equilibrium with $\ce{HF}$, as, $$\ce{H+ + F- <=> HF}$$ This equilibrium satisfies the equation $$\ce{\frac{[H^+][F^-]}{[HF]}= 10^{-3.17 } => \frac{10^{-3}(2S - x)}{x} = 10^{-3.17} ...(i)}$$.

And another equation which you have to solve is just regarding $K_{sp}$, i.e $$S(2S-x)^2 = 3.7 \times 10^{-11} .....(ii)$$ From first equation, you will get $\frac{2S}{x} = 1.676$ ; if you substitute for $x$ in the second equation by this, you can now solve for $S$ and the answer will be $S= 3.845 \times 10^{-4}$.

Thus , the factor will be $\frac{3.845 \times 10^{-4}}{2.1 \times 10^{-4}} = 1.83$

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Here was my attempt #$1$ at a solution.

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

Using the information at pH=7, I estimated that $K_{sp}\approx3.7\cdot10^{-11}$, which means that barely any HF formed.

This is sort of backwards. At pH=7 there is barely any HF formed, so $\ce{[F-] \approx 2[Ca^{2+}]}$, and thus $K_{sp}\approx3.7\cdot10^{-11}$

I assumed that only $HF$ forms and not $CaOH$ because it's insoluble.

Nice! But to quibble it should be $\ce{Ca(OH)2}$.

I got $\displaystyle \frac{[H^+][F^-]}{HF}=10^{-3.17}$

We have $ \displaystyle \sqrt{\frac{3.7\cdot10^{-11}}{[Ca^{2+}]}}$ fluoride ions.

We rewrite $K_a=10^{-3.17}=\displaystyle\frac{10^{-3}[F^-]}{2[Ca^{2+}]-F^-}$ and plug in the concentration of fluoride ions in terms of calcium ions.

Again a bit sloppy but all correct to this point. Id write out that $\ce{ 2[Ca^{2+}] = [F^-] +[HF]}$and thus $\ce{[HF] = 2[Ca^{2+}] - [F^-]}$

$$K_a = 10^{-3.17} =\dfrac{\ce{[H+][F-]}}{\ce{[HF]}} =\dfrac{(10^{-3})\ce{[F^-]}}{\ce{2[Ca^{2+}]-[F^-]}}\tag{1}$$

$$ 6.761\cdot10^{-4} =\dfrac{10^{-3}\ce{[F^-]}}{\ce{2[Ca^{2+}]-[F^-]}}\tag{2}$$

$$ 0.6761 =\dfrac{\ce{[F^-]}}{\ce{2[Ca^{2+}]-[F^-]}}\tag{3}$$

$$ \ce{2[Ca^{2+}]-[F^-]} =\dfrac{\ce{[F^-]}}{0.6761} = 1.4791\ce{[F^-]} \tag{4}$$

$$ \ce{[Ca^{2+}]} = 1.240\ce{[F^-]} \tag{5}$$

$$ \ce{[Ca^{2+}]} = 1.240\sqrt{\dfrac{3.7\cdot10^{-11}}{\ce{[Ca^{2+}]}}} \tag{6}$$

$$ \ce{[Ca^{2+}]^3} = (1.240)^2 \times 3.7\cdot10^{-11}\tag{7}$$

$$ \ce{[Ca^{2+}]} = \sqrt[3]{(1.240)^2 \times 3.7\cdot10^{-11}} = 3.846\cdot10^{-4} \tag{8}$$

So you had the right equations, you just flubbed the math.

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Soumik got the right answer, but the method seems weird. Let's do this a different way.

We have three chemical equations to use.

$$\ce{CaF2 <=> Ca^2+ + 2F-}\tag{1}$$

$$\ce{HF <=> H+ + F-}\tag{2}$$

$$\ce{[Ca^{2+}] = \frac{1}{2}([F-] + [HF])}\tag{3}$$

From equation (2) we know that

$$K_a = 10^{-3.17 } = 6.761\cdot10^{-4} = \ce{\frac{[H^+][F^-]}{[HF]}}\tag{5} $$

rearranging we get

$$\ce{[HF]} = \dfrac{\ce{[H^+][F^-]}}{K_a}\tag{6}$$

Substituting this expression for $\ce{[HF]}$ into equation (3) gives

$$\ce{[Ca^{2+}]} = \frac{1}{2}\left(\ce{[F-]}+ \dfrac{\ce{[H^+][F^-]}}{K_a}\right)= \frac{1}{2}\ce{[F-]\left(\dfrac{K_a+\ce{[H^+]}}{K_a} \right)}\tag{7}$$

rearranging we get

$$\ce{[F-]} = \ce{2[Ca^{2+}]}\left( \dfrac{K_a}{K_a+\ce{[H^+]}} \right)\tag{8} $$

From equation (1) we get

$$K_{sp} = \ce{[Ca^{2+}][F-]^2}\tag{9}$$

or substituting (8) and simplfying

$$\ce{[Ca^{2+}]} = \sqrt[3]{\dfrac{K_{sp}}{4\left( \dfrac{K_a}{K_a+\ce{[H^+]}} \right)}}\tag{10}$$

Now for pH=7 and pH=3 let's evaluate the term $4\left(\dfrac{K_a}{K_a+\ce{[H^+]}}\right)^2$.

$$\text{@pH=7}\quad4\left(\dfrac{K_a}{K_a+\ce{[H^+]}}\right)^2 = 4\left(\dfrac{6.761\cdot10^{-4}}{6.761\cdot10^{-4}+1\cdot10^{-7}}\right)^2 = 3.999\tag{11}$$

$$\text{@pH=3}\quad4\left(\dfrac{K_a}{K_a+\ce{[H^+]}}\right)^2 = 4\left(\dfrac{6.761\cdot10^{-4}}{6.761\cdot10^{-4}+1\cdot10^{-3}}\right)^2 = 0.6509\tag{12}$$

So

$$\text{@pH=7}\quad K_{sp}= 3.999\ce{[Ca^{2+}]^3} = 3.999(2.1\cdot10^{-4}) = 3.703\cdot10^{-11}\tag{13}$$

$$\text{@pH=3}\quad \ce{[Ca^{2+}]} = \sqrt[3]{\dfrac{K_{sp}}{0.6509}}=3.846\cdot10^{-4}\tag{14}$$

and finally

$$\text{ratio} = \dfrac{\ce{[Ca^{2+}]}_{\text{pH=3}}}{\ce{[Ca^{2+}]}_{\text{pH=7}}} = \dfrac{3.846\cdot10^{-4}}{2.1\cdot10^{-4}}=1.831 \ce{->[Rounding]} 1.8$$

NOTE - There are only two significant figures in all the data given, so the answer should be 1.8, not 1.83.

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  • $\begingroup$ i understand both of the methods, and they make perfect sense. so where did I make a mistake in my method? $\endgroup$ – Barry Chau Nov 14 '18 at 18:37

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