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. When pH is 7.8

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, pOH = 14 - 7 = 7. If pH = 11.5, then pOH =14.0 - 11.5 = 2.5. And if pH = 12.7, then pOH = 14.0 - 12.7 = 1.3. pOH = -Log[OH-], so if pOH = 7, then Log[OH-] = -7, and [OH-] = 1 x 10^-7. If pOH = 2.5, then Log[OH] = -2.5, and [OH-] = 3.2 x 10^-3 (calculator). If pOH = 1.3, then [OH-] = 5.0 x 10^-2. For pH 7: Ksp = (x)(1x10^-7)^2 = 7.9 x 10^-16 x = 7.9 x 10^-2 M = molar solubility. For the other two cases, solve the same expression, using 3.2 x 10^-3 and 5.0 x 10^-2 for [OH-]

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