Posted by Sarah (C) on Saturday, April 23, 2011 at 3:36pm.
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.
Related Questions
Chemistry - you add excess sodium sulfate to a solution of a soluble barium ...
Chemistry - you add excess sodium sulfate to a solution of a soluble barium ...
Chemistry - The reaction of aqueous solutions of barium chloride and sodium ...
chem - calculate the solubility of barium sulfate (Ksp=1.1x10^-10) in a) water b...
AP Chem - A chemist added an excess of sodium sulfate to a solution of a soluble...
Chemistry - Aluminum sulfate reacts with barium chloride to yield barium ...
AP Chem - Whew. long question. Barium carbonate is the source of barium ...
chemistry - how much of a 1.50M solution sodium sulfate solution in milliliters ...
chemistry - If you begin with 100.0 mL of a 3.0 Molar solution of ammonium ...
Chemistry - If you begin with 100.0 mL of a 3.0 Molar solution of ammonium ...
For Further Reading
