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. pH is 7.8, 11.5, and 12.2

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 = 7.8, pOH = 14 - 7.8 = 6.2. If pH = 11.5, then pOH =14.0 - 11.5 = 2.5. And if pH = 12.2, then pOH = 14.0 - 12.2 = 1.8. pOH = -Log[OH-], so if pOH = 6.2, then Log[OH-] = -6.2, and [OH-] = 6.31x 10^-7. If pOH = 2.5, then Log[OH] = -2.5, and [OH-] = 3.2 x 10^-3 (calculator). If pOH = 1.8, then [OH-] = 1.58 x 10^-2. For pH 7.8: Ksp = (x)(6.31x10^-7)^2 = 7.9 x 10^-16 x = 1.98 x 10^-3 M = molar solubility. For the other two cases, solve the same expression, using 3.2 x 10^-3 and 1.580 x 10^-2 for [OH-]

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