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Cr(OH) 3(s) Cr 3+ (aq) +3OH - (aq) K sp = 6.7*10 -31 Cr 3+ (aq) +4OH - (aq) Cr(OH) 4 - K f =

Cr(OH)3(s)<-> Cr3+(aq)+3OH-(aq) Ksp = 6.7*10-31
Cr3+(aq)+4OH-(aq)<->Cr(OH)4- Kf = 8*1029

The solubility of Cr(OH)3 in pure water is: 1.3 * 10-8 M

The solubility of Cr(OH)3 in 0.50M NaOH when the complex ion Cr(OH)4-is formed is:

a) 0.32 M

b) 0.59 M

c) 0.27 M

d) 0.18 M

I know the answer is D, but don't know the process ...thanks

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