Question

Solid ammonium phosphate is slowly added to 75.0 mL of a 0.199 M calcium iodide solution until the concentration of phosphate ion is 0.0609 M. The percent of calcium ion remaining in solution is _____ %.

Answer 1

(NH4)3PO4 (s) to V = 75 mL of CaI2

the next reaction will occur:

Ca+2 + PO4-3(aq) --> Ca3(PO4)2

Ksp = [Ca+2]^3[PO4-3]^2

Ksp = 2.07*10^-33

2.07*10^-33 = [Ca+2]^3[PO4-3]^2

if phosphate concentration is 0.0609

then

2.07*10^-33 = [Ca+2]^3(0.0609)^2

[Ca+2] = ((2.07*10^-33) /(0.0609)^2)^(1/3) = 8.2333*10^-11 M

initially --> 0.199 M

so

% = (8.2333*10^-11) / 0.199 * 100 = 4.13733*10^-8 %, meaning that most phosphate has precipitate along with the Calcium ions