Calculate the molar solubility of Fe(OH)2 in a buffer solution where the pH has

Question

Calculate the molar solubility of Fe(OH)2 in a buffer solution where the pH has been fixed at the indicated values. Ksp = 7.9 x 10-16.

(a) pH 7.2 __?__ M

Explanation / Answer

Fe(OH)2(s) ===> Fe2+(aq) + 2OH-(aq) Ksp = [Fe2+][OH-]^2 7.9 x 10^-16 Molar solubility of Fe(OH)2 = [Fe2+], so find [OH-] in each case, and solve for [Fe2+]. In each case, let [Fe2+] = x. pH + pOH = 14.0. So if pH = 7.2, pOH = 14 - 7.2 = 6.8 pOH = -Log[OH-], so if pOH = 6.8, then Log[OH-] = -6.8, and [OH-] = 1 x 10^-6.8 For pH 7: Ksp = (x)(1x10^-6.8)^2 = 7.9 x 10^-16 x = 17.73x 10^-2 M = molar solubility.

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