# A solution contains 1.43×10-2

A solution contains 1.43×10-2 Mpotassium bromide and1.37×10-2 M ammoniumchromate.Solid silver nitrate is added slowly to thismixture.A. What is the formula of thesubstance that precipitates first?

 formula =

B. What is the concentration ofsilver ion when thisprecipitation first begins?[Ag+] = M

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^-6M

AgBr will be form first precipitate.

Ionic products > KSp precipitate will form first

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

3.78*10^-11* 1.43×10-2

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

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