Why is this equation true for a salt in acid?

I previously asked a [question](https://chemistry.stackexchange.com/questions/164977/how-can-i-calculate-ksp-without-initial-values) about a calculation (below)



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> Calculate the molar solubility of $\ce{SrF2}$ in a solution buffered at pH = 2.00. ($K\_{a}$ for HF is $7.2 \times 10^{-4}$). The $K\_{sp}$ of $\ce{SrF2}$ is $K\_{sp} = 7.9 \times 10^{-10}$
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The answers used the fact that:


$$\ce{[HF] + [F-] = 2[Sr^2+]}$$


But I don't understand this. I can understand why this is true in a pure water solution:
$$\ce{[F-] = 2[Sr^2+]}$$


This must be true because if $\ce{SrF2}$ dissolves, it must release $\ce{2F-}$ ions for every $\ce{Sr^{+2}}$ ion it releases. This makes sense.


But for the case where there is a HF buffer, I don't get why you add on the $\ce{[HF]}$. What I personally expect is that because the $\ce{HF}$ will undergo this equilibrium:


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


then it will decrease the solubility of $\ce{SrF2}$ due to the $\ce{F-}$ being a common ion. But I really don't see how this translates to adding a $\ce{[HF]}$ onto the expression above.




It is an example for parallel reaction. The reactions taking place are $\ce{SrF2 <=> Sr^2+ +2F−}$ and $\ce{H+ +F− <=> HF}$. The $\ce{F−}$ ions formed from the first reaction reacts simultaneously with $\ce{H+}$ ions in the medium forming
$\ce{HF}$.


For your better understanding let us say that the second reaction is *freezed* for sometime. So $2[\ce{Sr^2+}]\_i = [\ce{F-}]\_i$. Now let us say that second reaction is *unfreezed*. So some quantity of fluoride ions gets converted to $\ce{HF}$. And from second reaction the lost amount of fluoride ions should be equal to the amount of $\ce{HF}$ formed. So $[\ce{F-}]\_i =[\ce{F-}]\_f + [\ce{HF}]$ and $[\ce{Sr^2+}]\_i =[\ce{Sr^2+}]\_f$. Here subscripts denote initial and final concentrations. So finally the relation between these concentrations is
$$2[\ce{Sr^2+}]=[\ce{F-}] + [\ce{HF}]$$


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