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Solubility products

Solubility products. CHEM 114 Fundamental Chemistry. If we could assume that activities were concentrations, we could use the solubility product to calculate the molar solubility of AgI. Ag + ( aq ) + I – ( aq ) ⇌ AgI ( s ). a Ag+ a I– = 8.5 × 10 –17 ~ [Ag+][I – ]/M 2.

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Solubility products

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  1. Solubility products CHEM 114 Fundamental Chemistry

  2. If we could assume that activities were concentrations, we could use the solubility product to calculate the molar solubility of AgI Ag+ (aq) + I– (aq) ⇌ AgI (s) aAg+aI– = 8.5 × 10–17 ~ [Ag+][I–]/M2 Solubility products and activities If x moles of AgI (s) dissolves, then [Ag+][I–] = x2; x = 9.2 ×10–9 M. So the solubility of AgI in water at 25° C is 9.2 ×10–9 M This is a good approximation in very dilute solution (< 10–6 M) and a fair approximation in semi-dilute solution (< 10–4 M). To be accurate it needs adjustment for the activity coefficient : aAg+ = γAg+ [Ag+] For water at 298 K Aγ = 1.17 mol–1/2 kg1/2 uncharged solute I = ½ ΣZi2mi As I increases, ln γ becomes more negative, and γ decreases CHEM 114 Fundamental Chemistry ionic solute

  3. 2Ag+ (aq) + CrO42– (aq) ⇌ Ag2CrO4 (s) Ksp = aAg+2aCrO42– = 1.1 × 10–12 @ 25°C 4x3 = 1.1 × 10–12 so x = 6.5 × 10–5 Common ion effect CHEM 114 Fundamental Chemistry

  4. Al3+ (aq) + 3 OH– (aq) ⇌ AI(OH)3 (s) Ksp = aAl3+aOH–3 = 1.3 × 10–33 @ 25°C Solubility and pH Instead of calculating the molar solubility, let’s calculate the solubility of Al3+ at pH 7 and pH 3 aAl3+aOH–3 = 1.3 × 10–33 aAl3+ = 1.3 × 10–33/aOH–3 At pH 7, aOH– = 10–7 aAl3+ = 1.3 × 10–33/10–21 = 1.3 × 10–12 At pH 3, pOH = 11, aOH– = 10–11 aAl3+ = 1.3 × 10–33/10–33 = 1.3 Aluminum ions are incredibly insoluble at pH 7, but very soluble at pH 3. • Pb2+(aq) + 2 OH– (aq) ⇌ Pb(OH)2 (s) Ksp = aPb2+aOH–2 = 1.43 × 10–20 @ 25°C At pH 8, aOH– = 10–6 aPb2+ = 1.43 × 10–20/10–12 = 1.43 × 10–8 CHEM 114 Fundamental Chemistry At pH 6, aOH– = 10–8 aPb2+ = 1.43 × 10–20/10–16 = 1.43 × 10–4

  5. Question: what happens when we add 1 mol of Hg(CH3COO)2 to 1 L of water? Precipitation equilibria and pH • Hg(CH3COO)2 (s) → Hg2+ (aq) + 2 CH3COO– (aq) • CH3COO– (aq) + H2O → CH3COOH (aq) + OH– (aq) pKa = 4.74 ⇒ pKb = 9.26 Kb= 5.50 ×10–10 x = [OH–] = 3.3 ×10–5 x2/(2 – x) = 5.50 × 10–10 • Hg(OH)2 (s) ⇌ Hg2+ (aq) + 2 OH– (aq) • But Hg(OH)2 has a small solubility product! Qsp>> Ksp Ksp= 3.0 ×10–26 Qsp= 1 × (3.3 ×10–5) = 1.1 ×10–9 • So if we try to dissolve mercury (II) acetate in ordinary water, Hg(OH)2 will precipitate! CHEM 114 CHEM 114 Fundamental Chemistry • Hg(OH)2 decomposes to give H2O + HgO, which is orange.

  6. The same will happen with any salt of a weak acid if the hydroxide is very insoluble. Why you can’t buy real aluminum acetate... CHEM 114 CHEM 114 Fundamental Chemistry

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