Calculate the molar solubility of Cr(OH)3 in 0.32 M NaOH.

Question

for Cr(OH)4-, Kf = 8 x 10^(29)

for Cr(OH)3, Ksp = 6.7 x 10^(-31)

the answer should be expressed using one significant figure.

0.2 and 0.4 are incorrect answers (already tried them)

Cr(OH)3 is an amphoteric hydroxide

Cr(OH)4- is a complex ion

Answer 1

Cr(OH)3(s)  <- -> Cr+++  +  3OH-

___/____________S______0.32*beta__

beta = 1 / (1 + Kf*[OH-]^4) = 1 / (1 + 8*10^29*(0.32)^4) = 1.19*10^-28

so

Kps = 6.7*10^-31 = S*(0.32*1.19*10^-28)^3

S = 1.20*10ì55  => completely soluble