The molar solubility of M(OH)3 in 0.4 M MX3 solution in terms of solubility product of M(OH)3 is

Following equilibria gets simultaneous establishedM(OH)3(s)⇌M(OHS)3(aq)→M+3S+0.4(aq)+3OH-3s(aq)And the solubility of M(OH)3 will get suppressed in presence of 0.4 M MX3 solution.MX3→M0.4+S'+3(aq)+3X3×0.4-(aq)the solubility of M(OH)3 is suppressed, S' will be negligible. i.e. 0.4 + S' ~ 0.4Ksp = (S' + 0.4)  x (3S')3⇒  Ksp = 0.4 x S'3 x 27⇒Ksp10.8=S'3⇒S'=Ksp10.813