# Why is this equation true for a salt in acid?

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.

• 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. May 13 at 6:47
• 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-. May 13 at 12:55

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 $$2[\ce{Sr^2+}]=[\ce{F-}] + [\ce{HF}]$$
• (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: $[\ce{F-}]_f =[\ce{F-}]_i - [\ce{HF}]$ May 14 at 12:38
• (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 May 14 at 12:48
• @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 $\ce{K_{sp}=[\ce{Sr^2+}][\ce{F-}]^2}$ May 14 at 13:47