Find the concentration of Ca2+, in equilibrium with CaSO4(s) and 0.035 M SO4-2.
I am using Ksp from the web of 4.93E-5. Be sure to use the value for Ksp of CaSO4 in your text or in your notes.
...................CaSO4 ===> Ca^2+ + SO4^2-
I.....................solid...............0...........0
change...........solid..............x.............x
E......................solid..............x............x
Ksp = 4.93E-5 = (Ca^2+)(SO4^2-)
You know Ksp. The problem tells you [SO4]^2- is 0.035 M
Plug in those values and solve for (Ca^2+) and you're done. That wasn't bad was it.
Thank you so much for your help.
To find the concentration of Ca2+ in equilibrium with CaSO4(s) and 0.035 M SO4^2-, you need to use the solubility product constant (Ksp) expression for CaSO4 and the stoichiometry of the dissociation reaction.
The dissociation of CaSO4 in water can be represented as follows:
CaSO4(s) ⇌ Ca2+(aq) + SO4^2-(aq)
The Ksp expression for this reaction is:
Ksp = [Ca2+][SO4^2-]
We know that the concentration of SO4^2- is 0.035 M. To find the concentration of Ca2+, we need to know the value of Ksp.
According to the literature, the Ksp value for CaSO4 is 4.93 x 10^-5 at 25°C.
Using the Ksp expression and the given Ksp value, we can set up the equation as follows:
4.93 x 10^-5 = [Ca2+] * (0.035)
To solve for [Ca2+], divide both sides of the equation by 0.035:
[Ca2+] = (4.93 x 10^-5) / (0.035)
Evaluate the expression to find the concentration of Ca2+ in equilibrium with CaSO4(s) and 0.035 M SO4^2-.