Question

A solution contains 1.43×10^-2 M potassium bromide and 1.37×10^-2 M ammonium chromate.
Solid silver nitrate is added slowly to this mixture.

A. What is the formula of the substance that precipitates first?

formula =

B. What is the concentration of silver ion when this precipitation first begins?
[Ag⁺] = M

Answer 1

KBr(aq) + AgNO3 (aq) ----------------> AgBr(s) + KNO3(aq)

KBr -------------> K^+ (aq) + Br^-

1.43*10^-2M                           1.43*10^-2

Ksp of AgBr   = 5.4*10^-13

AgBr -------------> Ag^+ (aq) + Br^- (aq)

   Ksp   = [Ag^+][Br^-]

5.4*10^-13   = [Ag^+] *1.43*10^-2

[Ag^+]     = 5.4*10^-13/1.43*10^-2   = 3.78*10^-11 M

(NH4)2CrO4 (aq) + 2AgNo3(aq) -----------------> Ag2CrO4(s) + 2NH4NO3(aq)

(NH4)2CrO4 --------------------> 2NH4^+ (aq) + CrO4^2-

1.37*10^-2M                                                        1.37*10^-2M

ksp of Ag2CrO4 = 1.12*10^-12

      Ag2CrO4(s) ----------------------> 2Ag^+ (aq) + CrO4^2- (aq)

        Ksp   = [Ag^+]^2[CrO4^2-]

     1.12*10^-12 = [Ag^+]^2 *1.37*10^-2

     [Ag^+]^2    = 1.12*10^-12/1.37*10^-2   = 8.17*10^-11

    [Ag^+]     = 9*10^-6 M

AgBr will be form first precipitate.

Ionic products > KSp precipitate will form first

[Ag^+][Br^-] >Ksp

3.78*10^-11* 1.43×10^-2

AgBr>>>>>answer

the concentration of silver ion precipitation first begins
[Ag⁺] = 3.78*10^-11 M >>>>>answer