A) Solubility of BaSO4 is worked similar to the CaCO3 problem. The principle difference is that the solution has Na2SO4 in it. Note Na2SO4 is a common ion (the sulfate ion is the common ion). What does that do. Sulfate comes from two sources; i.e., from BaSO4 (where it will be x) and from Na2SO4 (where it will be 0.05M). So the BaSO4 problem will be
Ksp = (Ba^2+)(SO4^2-)
Ksp = (x)(x+0.05) and solve for x.
B)I assume X is a halogen which then means M must be +3 element.
MX3(s) ==> M^3+ + 3X^-
Ksp = (M^3+)(X^-)^3
So set up the ICE chart in terms of s (follow the examples before where we let solubility be x) and plug that into Ksp expression. The correct answer is listed in the choices.
what is the solubility of barium sulfate in a solution containing 0.050 M sodium selfate? The Ksp value for barium sulfate is 1.1E-10.
An insoluble salt with formula MX3 has a solubility product constant written in terms of solubility,s, in the form
a) Ksp=s2
b)Ksp=4s3
c) Ksp=27s4
d)Ksp=256s5
1 answer
1 answer