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I previously asked a question about a calculation (below)

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 $K_{sp} = 7.9 \times 10^{-10}$

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.

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  • $\begingroup$ The link is broken, this should be OK. But AFAIK, there is nothing said about HF buffer. In fact, nothing explicit was said about the pH buffer but pH value. $\endgroup$
    – Poutnik
    May 13 at 6:47
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    $\begingroup$ The expression in a solution buffered at pH = 2.00. (Ka for HF is 7.2×10−4). says that 1/ there is a pH buffer with pH=2.00 AND 2/ the acid HF has acidity constant K_a=7.2×10−4. It does not say the pH buffer is based on HF/F-. $\endgroup$
    – Poutnik
    May 13 at 12:55

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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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  • $\begingroup$ (1 of 2) Sorry but I don't get it. If there is some $\ce{F-}$ being turned into HF, then shouldn't there be a decrease in the amount of $\ce{F-}$? Thus we should instead write: $[\ce{F-}]_f =[\ce{F-}]_i - [\ce{HF}]$ $\endgroup$
    – John Hon
    May 14 at 12:38
  • $\begingroup$ (2 of 2) Unless you mean that the SrF2 would disassociate more to make up for it? But then this line wouldn't be true: $[\ce{Sr^2+}]_i =[\ce{Sr^2+}]_f$ since we have changed the amount of F-. Hence the $[\ce{Sr^2+}]$ will be different too as Ksp is constant $\endgroup$
    – John Hon
    May 14 at 12:48
  • $\begingroup$ Sorry now I have edited $\endgroup$
    – Infinite
    May 14 at 13:39
  • $\begingroup$ @JohnHon Let us say that quantity of $\ce{Sr^2+}$ formed is more than the solubility obtained from $\ce{K_{sp}}$ equation. Now the flouride ion concentration decreases in such a way that $\ce{K_{sp}=[\ce{Sr^2+}][\ce{F-}]^2}$ $\endgroup$
    – Infinite
    May 14 at 13:47
  • $\begingroup$ It is just a way of interpretation. In reality no reaction freezes for sometime and proceeds after sometime. $\endgroup$
    – Infinite
    May 14 at 13:51

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