19.3 Equilibria of Slightly Soluble Ionic Example: Write the solubility-product Compounds expression for Al(OH)3. – Slightly soluble ionic solids reach equilibrium with their → The subscripts in the formula become powers in saturated solutions at very low concentrations of the the solubility-product expression dissolved solute 3+ - 3 –The dissolved portion of the ionic solid is assumed to be ⇒ Ksp = [Al ][OH ] completely dissociated to ions (not always the case) ¾ Sulfides are slightly different since S2- acts as a - The Solubility-Product Constant (Ksp) strong base in water and converts entirely to OH ¾For an ionic solid, MpAq, in equilibrium with its Example: Ag2S saturated solution, the equilibrium constant of the Ag S(s) ↔ 2Ag+ + S2- dissolution process is called solubility-product 2 S2- + H O(l) → HS- + OH- M A (s) ↔ pMm+ + qAn- 2 p q + - - + → Ag2S(s) + H2O(l) ↔ 2Ag + HS + OH m+ p n- q + 2 - - K sp = [M ] [A ] ⇒ Ksp = [Ag ] [HS ][OH ] • Ksp is a measure of the extent to which the Example: The solubility of Pb(IO3)2 is 0.022 g/L solubility equilibrium proceeds to the right at 25°C. Calculate Ksp of Pb(IO3)2. → Convert the solubility to molar solubility, s ↑Ksp ⇔↑Solubility – K depends on temperature g Pb(IO ) 1 mol Pb(IO ) -5 mol sp s = 0.022 3 2 × 3 2 = 3.9×10 L 557 g Pb(IO3 )2 L Calculations Involving Ksp → Express K through the molar solubility, s • Molar solubility (s) – the molarity of the sp 2+ - saturated solution (mol/L) Pb(IO3)2(s) ↔ Pb + 2IO3 2+ – The solubility is often expressed in g/L or 1 mol Pb(IO3)2 → 1 mol Pb - grams/100 mL of solution → can be easily 1 mol Pb(IO3)2 → 2 mol IO3 converted to mol/L 2+ - ⇒ [Pb ] = s and [IO3 ] = 2s – Ksp can be experimentally determined by 2+ - 2 2 3 measuring the molar solubility, s ⇒ Ksp = [Pb ][IO3 ] = s(2s) = 4s -5 3 -13 – s can be calculated if Ksp is known ⇒ Ksp = 4×(3.9×10 ) = 2.5×10 1 -5 Example: The Ksp of Ag2SO4 is 1.4×10 at ¾Ksp can be used as a guide in comparing molar solubilities of different ionic compounds (works 24°C. Calculate the molar solubility of Ag2SO4. → Express K through the molar solubility, s only if the compounds have the same number of ions sp in the formula) + 2- Ag2SO4(s) ↔ 2Ag + SO4 ↑Ksp ⇔↑s + 2- ⇒ [Ag ] = 2s and [SO4 ] = s + 2 2- 2 3 ⇒ Ksp = [Ag ] [SO4 ]= (2s) s = 4s 3 1/3 ⇒ s = Ksp/4 ⇒ s = (Ksp/4) ⇒ s = (1.4×10-5/4)1/3 = 1.5×10-2 M + 2- Note: The problem Ag2SO4(s) ↔ 2Ag + SO4 can be solved using an i excess 0 0 ice table: c -s +2s +s 2 3 ⇒ Ksp = (2s) s = 4s e excess 2s s The Common Ion Effect Example: At a given temperature, Ksp of AgBr is ¾ For a slightly soluble ionic solid, MA 7.7×10-13. Estimate the solubility of AgBr in: n+ n- n+ n- a) Pure H O b) 0.10 M CaBr solution MA(s) ↔ M + A Ksp = [M ][A ] 2 2 –If Mn+ is added by means of the soluble salt a) In pure H2O (MB), the equilibrium shifts to the left toward → Express Ksp through the molar solubility, s formation of more MA(s) AgBr(s) ↔ Ag+ + Br- n+ – MA and MB have a common ion (M ) i excess 0 0 – Similarly, if An- is added to the solution, the c -s +s +s equilibrium also shifts to the left toward e excess s s formation of more MA(s) ⇒ [Ag+] = s and [Br-] = s ⇒The molar solubility decreases in the presence ⇒ K = [Ag+][Br-]= s2 of common ions sp 1/2 -13 1/2 -7 ⇒ s = (Ksp) = (7.7×10 ) = 8.8×10 M 2 b) In 0.10 M CaBr2 solution The Effect of pH on Solubility - - 0.10 M CaBr2 → 0.20 M Br (Br is the common ion) ¾ The solubility of some ionic solids in water is → Express Ksp through the molar solubility, s greatly affected by the pH + - AgBr(s) ↔ Ag + Br – Metal hydroxides (Fe(OH)3, Mg(OH)2, …) – the - + i excess 0 0.20 anion (OH ) reacts with added H3O c -s +s +s ⇒The solubility can be improved by adding acids e excess s 0.20 + s Example: Mg(OH)2 ⇒ K = [Ag+][Br-]= s(0.20 + s) 2+ - sp Mg(OH)2(s) ↔ Mg + 2OH Assume s << 0.20 K = s(0.20) + → ⇒ sp → If acid (H3O ) is added: -13 -12 + - + - ⇒ s = Ksp/0.20 = 7.7×10 /0.20 = 3.8×10 M →H3O consumes OH (H3O + OH → 2H2O) → Check assumption → OK →The equilibrium shifts to the right and the solubility increases ⇒ The solubility is much lower in 0.1 M CaBr2 – Salts of weak acids (CaCO3, BaSO3, PbF2, ZnS, Predicting Precipitation – Q versus K …) – the anion of the salt is a weak base which sp sp + ¾ The reaction quotient of the dissolution reacts with added H3O ⇒The solubility can be improved by adding acids process (Qsp) is defined in the same way as K and at equilibrium Q = K Example: BaSO (a salt of H SO ) sp sp sp 3 2 3 ¾ To predict precipitation, compare Q to K 2+ 2- sp sp BaSO3(s) ↔ Ba + SO3 → If Q > K , precipitation occurs + sp sp → If acid (H3O ) is added: → If Qsp < Ksp, dissolution occurs (no precipitation) 2- + - SO3 + H3O → HSO3 + H2O(l) → If Qsp = Ksp, no change occurs →H O+ consumes SO 2- 3 3 Example: Does a precipitate form after →The equilibrium shifts to the right and the -4 solubility increases mixing of 200. mL 1.0×10 M AgNO3 with -6 Example: CaCO (s) ↔ Ca2+ + CO 2- 900. mL 1.0×10 KCl? 3 3 -10 2- + (K = 1.8×10 for AgCl) CO3 + 2H3O → H2CO3 +H2O(l) → CO2(g) + 2H2O(l) sp 3 (aq) (aq) (s) (aq) → AgNO3 + KCl → AgCl + KNO3 → Calculate Qsp: + - + - -5 -7 → Net ionic equation: Ag + Cl → AgCl(s) Qsp = [Ag ][Cl ] = (1.8×10 )×(8.2×10 ) + - -11 → Reverse: AgCl(s) → Ag + Cl Qsp =1.5×10 -10 + - -10 Ksp = 1.8×10 Ksp = [Ag ][Cl ]= 1.8×10 Vtot = 0.200 + 0.900 = 1.100 L ⇒ Qsp < Ksp → Initial concentrations after mixing: ⇒ No precipitation occurs -4 mol 1.0×10 × 0.200 L mol [Ag+ ] = L = 1.8×10-5 1.100 L L -6 mol 1.0×10 × 0.900 L mol [Cl- ] = L = 8.2×10-7 1.100 L L 4.