Solid chromium (III) hydroxide (Cr(OH)3) is added to pure water so that at equil

Question

Solid chromium (III) hydroxide (Cr(OH)3) is added to pure water so that at equilibrium some solid remains undissolved. Given that the solubility product is 6 x 10-31 M4, what is the equilibrium concentration of the hydroxide ion (OH-) in the water? Please show formulas used, your answer with units and calculations as text below or an attached file.

Explanation / Answer

Cr(OH)3 ------------> Cr+3 + 3OH- ------------------------X------3X+10^-7 Ksp = 6*10^-31 = X * (3X+10^-7)^3 neglect 3X we get 6 * 10^-31 = X * 10^-21 => X = 6 * 10^-10 [OH-] = 3X + 10^-7 = 1.018 * 10^-7 M

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