Chemistry 212
posted by Pookie .
What is the solubility of RaSO4 in 0.1381M Na2SO4? Ksp = 4.3x10^11?

Chemistry 212 
DrBob222
RaSO4 ==> Ra^+2 + SO4^2
Ksp = (Ra^+2)(SO4^2).
If you let S = solubility RaSO4, then
Ra^+2 = S
SO4^2 = S from RaSO4 and 0.1381 M in Na2SO4 so a total of (S + 0.1381)
Substitute those values into Ksp expression and solve for S. 
Chemistry 212 
Pookie
Ksp = [Ra^+2][SO4^2]
4.3x10^11 = [x][x]
x=6.56x10^6
6.56x10^6 + .1381??
is that how im suppose to do it? 
Chemistry 212 
DrBob222
No.
4.3 x 10^11 = (x)(x+0.1381)
solve for x.
NOTE:
If you solve the equation as I wrote it you will need to solve a quadratic. Those are usually easy to do with modern calculators; however, you CAN make a simplifying assumption. You make the assumption that x + 0.1381 = 0.1381 (that is, that x is so small in comparison to 0.1381 that when added it isn't different than 0.1381). The equation then would look like this.
4.3 x 10^11 = (x)(0.1381) and solve for x. When you finish, then look at x+0.1381 and see how far it is from 0.1381. If x is so small that x+0.1381 is just 0.1381, then the assumption is ok. If not, the assumption is not ok and you must solve the quadratic. I ALWAYS make the assumption first and solve the easy equation, then check at the end and see if the assumption is ok. 
Chemistry 212 
Pookie
thanks a bunch!
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