Question

13.

A)Solid zinc acetate is slowly added to 75.0 mL of a 0.0683 M ammonium phosphate solution. The concentration of zinc ion required to just initiate precipitation is?

B)Solid potassium chloride is slowly added to 50.0 mL of a 0.0521 M silver acetate solution. The concentration of chloride ion required to just initiate precipitation is?

Answer 1

a)

When we add zinc acetate in ammonium phosphate the precipitation of zinc phosphate

Zn3(PO4)2 <-> 3Zn+2 2PO42-

Let the concentration of zinc ion needed for precipitation be x

Q = [Zn+2]³[PO4-]² = Ksp

then Q = Ksp = 9.1*10^-33 (csudh)

Concentration of Phosphate = concentration of ammonium phosphate = 0.0683 M

So,

x³*0.0683² = 9.1*10^-33

x = (9.1*10^-33/0.0683²)^1/3 M

x = 1.25*10^-10 M

so, we need concentration of Zn+2 = x = 1.25*10^-10 M

b)

When we add potassium chloride in silver acetate the precipitation of silver chloride

AgCl <-> Ag+ Cl-

Let the concentration of chloride ion needed for precipitation be x

Q = [Ag+][Cl-] = Ksp

then Q = Ksp = 1.8*10^-10 (ncsu)

Concentration of Phosphate = concentration of ammonium phosphate = 0.0521 M

So,

0.0521*x = 1.8*10^-10

x = (1.8*10^-10/0.0521) M

x = 3.45*10^-9 M

so, we need concentration of Cl- = x = 3.45*10^-9 M