molar solubility of CaF2 at pH3 given molar solubility at pH7 and pKa of HF 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?
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, $$\ce{CaF2 &lt;=&gt; Ca^2+ + 2F-}$$ $\ce{[Ca^2+ ] = S }$ mol/L. and $\ce{[F-] = 2S -}$ $x$ mol/L. ,as, this $x$ mol/L. of $\ce{F-}$ is in equilibrium with $\ce{HF}$, as, $$\ce{H+ + F- &lt;=&gt; HF}$$ This equilibrium satisfies the equation $$\ce{\frac{[H^+][F^-]}{[HF]}= 10^{-3.17 }   =&gt; \frac{10^{-3}(2S - x)}{x} = 10^{-3.17}  ...(i)}$$.

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 $\frac{3.845 \times 10^{-4}}{2.1 \times 10^{-4}} = 1.83$