Question
In: Chemistry
How many grams of CaF2 will be soluble in 100 ml of 0.250 M Ca(NO3)2? (Ksp...
How many grams of CaF2 will be soluble in 100 ml of 0.250 M Ca(NO3)2? (Ksp of CaF2 = 3.90*10^-11)
Solutions
Expert Solution
Ca(NO3)2(aq) -------------> Ca^2+ (aq) + 2NO3^- (aq)
0.25M 0.25M
CaF2(s) ----------------> Ca^2+ (aq) + 2F^- (aq)
x 2x+0.25
Ksp = [Ca^2+][F^-]^2
3.9*10^-11 = x*(2x+0.25)^2
3.9*10^-11 = x*(0.25)^2 [ 2x + 0.25 =0.25 , 2x<<<0.25]
x = 3.9*10^-11/(0.25)^2 = 6.24*10^-10
solubility of CaF2 = 6.24*10^-10 M
no of moles of CaF2 = molarity * volume in L
= 6.24*10^-10 *0.1 = 6.24*10^-11moles
mass of CaF2 = no of moles * gram molar mas
= 6.24*10^-11 *78 = 4.87*10^-9 g of CaF2
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