How about c) btw? I will repost it because I see that i didn't complete it.
c) Calculate the solubility in moles/L of Ag2SO4(s) in 0,22 moles/L AgNo3
My answer
1,6 * 10-5 (moles/L)3= (0,22 moles/L + 2x)2 * x
We assume that 2x is so small compared to 0,22 moles/L, that we remove it from the equation.
0,0484x = 1,6 * 10-5 (moles/L)3
x = 3,3 * 10-4 (moles/L)3
What I don't understand
My book's answer is the same, but in moles/L (3,3 * 10-4 moles/L) and not in (moles/L)3 (3,3 * 10-4 moles/L)3
Does anybody know why this is?