Chemistry
posted by Cat .
At 25C only .0190 mol of the generic salt AB2 is soluble in 1.00L of water. What is the Ksp of the salt at 25C?
AB2<>A^2+ + 2B^
I took the .0190 mol and doubled it and multiplied by .019 again and got 6.859x10^6 and got it wrong what should I do?
Please help Dr. Bob

Did you square the 0.019*2 for B^?
........AB2 ==> A^2+ + 2B^
I......solid....0.......0
C.....0.0190..0.019..0.019*2
E......solid...0.019..0.038
Ksp = (A^2+)(B^)^2
Ksp = (0.019)(0.038)^2
Ksp = 2.74E5 for the answer. I do it slightly different because it's faster on the calculator.
Ksp = (A^2+)(B^)^2
Kso = (x)(2x)^2 = 4x^3
Ksp = 4*(0.019)^3 = 2.74E5.
The usual student mistake is to double the 0.019 to find B^ but forget it is squared.