Unit 7: Topic 13 - pH and Solubility

Question 1

Justify why CaF₂ becomes more soluble when the pH is decreased using Le Châtelier’s principle.
(it’s assumed here that the acid we’re using to decrease the pH simply provides H+ ions and doesn’t interact with Ca²⁺ ions)

Answer 1

Firstly, we know that the dissolution equation for CaF₂ is :
CaF₂(s) <-> Ca²⁺(aq) + 2F⁻(aq)

This equation doesn’t explain why pH would effect Ksp so we have to dive deeper :

Notice that F⁻ is a weak base (conjugate base of HF), therefore we can write this equilibrium :

F⁻(aq) + H⁺(aq) <-> HF(aq), so in the presence of H⁺, HF will be made from F⁻

Let’s think about the consequences of decreasing pH, this means that H⁺ concentrations will increase. Therefore,
F⁻(aq) + H⁺(aq) <-> HF(aq) will shift to the right.

When pH is decreased, more F⁻ will be converted to HF. This means that the concentration of F⁻ will decrease.

If [F⁻] decreases then CaF₂(s) <-> Ca²⁺(aq) + 2F⁻(aq) will shift to the right to bring [F⁻] levels back up

But F⁻ must come from CaF₂ so to increase [F⁻] means to decrease the amount of CaF₂(s)

If we just decreased the amount of CaF₂(s), then more of it must have dissolved and so Ksp increases