Question

hat is the equilibrium Ag⁺ concentration when 3.25 L of a 0.102 M silver acetate solution are mixed with 3.41 L of a 0.179 M sodium bromide solution?

[Ag⁺] =  M

Answer 1

Calculate all initial concentrations:

Vtotal = 3.25 + 3.41 = 6.66 L

[Acetate] = M1*V1/Vtotal = (3.25*0.102)/(3.25+3.41) = 0.04977

[Ag+]1 = M1*V1/Vtotal = (3.25*0.102)/(3.25+3.41) = 0.04977

[Na+] = M2*V2/Vtotal = (3.41*0.179)/(3.25+3.41) = 0.091650

[Br-] = M2*V2/Vtotal = (3.41*0.179)/(3.25+3.41) = 0.091650

note that there will be precipitation

there is excess Br- so assume Br-

Ksp = [Ag+][Br-]

[Br-] = (0.091650-0.04977) = 0.04188

Ksp = 5.35×10–13

5.35*10^-13 =[Ag+](0.04188)

[Ag+]= (5.35*10^-13)/(0.04188) = 1.2774*10^-11 M

}