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.

(c) pH 12.6 __?__ 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 = 12.6, pOH = 14 - 12.6 = 1.4 pOH = -Log[OH-], so if pOH = 1.4, then Log[OH-] = -1.4, and [OH-] = 1 x 10^-1.4 For pH 12.6: Ksp = (x)(1x10^-1.4)^2 = 7.9 x 10^-16 x = 7.06 * 10^-7 M = molar solubility.

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