Ksp Solubility and pH

Question

At some other temperature, a solid sample of Ca(OH)2 is shaken with 0.0100 M CaCl2.

Once equilibrated, some solid Ca(OH)2 remains undissolved. The solution is filtered and a 15.00 mL sample requires 23.76 mL of 0.0100 M HCl to neutralize it.

a) Calculate the value for Ksp of Ca(OH)2 from this data.

The reaction of the solid can be written as

Ca(OH)2 ➡️ Ca^2+ + 2(OH)

We can figured the amount of moles of Ca(OH)2 by multiplying the volume and molarity of HCl. Then after we figure out the moles we can calculate the molarity of Ca(OH)2 by dividing (0.015L).

Now my question is when you’re calculating the concentration of Ca^2+ to put into the Ksp expression, do you need to account the 0.0100 M from CaCl2?

ksp= (Ca^2+)(2 * OH-)^2

I know that when CaCl2 was initially added to the solid, it caused a common ion effect thus pushing the equilibrium to the left towards Ca(OH)2. After this once the equilibrium is established, I was thinking the concentration of Ca^2+ would be the same as Ca(OH)2 that we figured from HCl? Hope that makes sense!

Answer 1

You would indeed have to consider the initial concentration of CaCl₂ when you write out a K_sp relationship for Ca(OH)_2. The value for [Ca²⁺] in the K_sp equation is going to be the initial concentration of Ca²⁺ in the solution from dissolved CaCl₂ (0.0100 M) plus the added concentration from whatever amount of Ca(OH)₂ dissolved.