Relationship b/t Qsp and Ksp?

Question

If a solution is 1 x 10^-5 M in Mn(NO3)2 and 1.5 x 10^-3 M in aqueous ammonia, will Mn(OH)2 precipitate?

Kb for NH3= 1.8 x 10^ -5

Ksp for Mn(OH)2 = 2.0 x 10^ -13

well, in the solution it said yes, b/c Qsp > Ksp

but I dont get it..how do you find Q?

Answer 1

Qsp = [Mn+2][[OH-]^2

[Mn+2] = 1x10^-5 M

NH3 + H2O --> NH4^+1 + OH-

Kb = [NH4^+1][OH-] / [NH3]

let X = [NH4+] = [OH-]

Kb = 1.8x10^-5 = X^2 / (0.0015 - X)

assume X << 0.0015

X^2 = 2.7x10^-8

X = [OH-] = 0.000164

Qsp = 1x10^-5 *(0.000164)^2

Qsp = 1x10^-5 * 2.7x10^-8

Qsp = 2.7x10^-13

Qsp > Ksp so a precipitate will form