How many grams of Ag2SO4 (formula weight = 312 g/mol) will dissolve in 1.5 L of water (Ksp = 1.5 × 10-5)?
How many grams of Ag2SO4 (formula weight = 312 g/mol) will dissolve in 1.5 L of water (Ksp = 1.5 × 10-5)?
The answer is 7.3 but i can't get the work, i know i am over thinking the problem
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Start with the usual chemical formula
Ag2SO4 == 2Ag + SO4
Let the molar solubility = x
x= [SO4]; so [Ag]= 2x
Set up a Ksp expression
Ksp= 1.5 x 10^-5= (x)(2x)^2
Simplify it--> 1.5 x 10^-5= 4x^3 Now use basic algebra. Divide the other side by 4, then find the cube root to solve for x, which is your molar solubility.
x= 1.55 x 10^-2 mol/L
Now convert mol/L to g/L by multiplying by the molar mass: 312g
(1.55 x 10^-2 mol / 1 L) x (312g / 1 mol) = 4.85g
This is where it gets tricky. 4.85g is how much would dissolve in ONE liter of water. The question asks for 1.5L (typical Ohio State chem for you.) But it's easy, just divide 4.85 by 2 and add that to the original 4.85g.
4.85/2= 2.42
4.85 + 2.42 = about 7.27 = 7.3g
I'm only assuming you're in Chem 123 at Ohio State. If that's the case, there's actually a problem in in the book that models it nicely. It's number 17.54 A. Check it out in the solutions posted on Carmen! Good luck, hope this helps!