Solubility product and Common Ion Effect

Very useful analytical approach for qualitative analysis of various acidic and basic radicals in their solution.

1. Solubility product and Common Ion effect Dr. Arun Sharma M.Sc. (Gold medalist), Ph.D . (DRDE) CSIR-UGC-NET-JRF, RPSC-SET

2. Solubility Product The solubility product of sparingly soluble electrolyte is equal to the product of ionic concentration in a saturated solution at the given temperature. Ionic compounds tend to be soluble in water. Because water is a polar solvent and stabilises the separated ions

3. Ksp, the solubility-product constant An equilibrium can exist between a partially soluble substance and its solution: AB(s) ⇄ A + (aq) + B - (aq) In saturated solutions dynamic equilibrium exists between undissolved solids and ionic species in solutions. Solids continue to dissolve and ion-pairs continue to form solids. At equilibrium condition, the rate of dissolution process is equal to the rate of precipitation.

4. The Solubility Product Expression AB(s) ⇄ A + (aq) + B - (aq) Solubility s s Solubility product, K sp = [A + ] [B - ] = s . s K sp = s 2 And s = [K sp ] 1/2 Solubility is expressed in terms of mol/Lit. and for soluble electrolyte it is expressed in terms of gm per 100 gm of the solvent. There are three conditions: [A + ] [B - ]   < K sp Unsaturated solution [A + ] [B - ]   =  K sp Saturate solution [A + ] [B - ]   >  K sp Oversaturated solution (Precipitation) K sp is known as the SOLUBILITY PRODUCT

5. Solubility & Solubility products AgCl (s) ⇌ Ag + (aq) + Cl - (aq) What is the equilibrium constant? K sp = [Ag + ] [ Cl - ] Now what about the value of the K sp ? K sp = 1.6 x 10 -10 Can we calculate the solubility, s? s s = s 2 s = (K sp ) 1/2 = 1.3 x 10 -5 mol/L For uni-univalent type of electrolytes

6. Solubility and Solubility Products Ag 2 CrO 4 (s) ⇌ 2Ag + (aq) + CrO 4 2- (aq) K sp = [Ag + ] 2 [CrO 4 2- ] = 9.0 x 10 -12 If s is the solubility of Ag 2 CrO 4 , then: [Ag + ] = 2 s and [CrO 4 2- ] = s K sp = (2s) 2 ( s ) = 4 s 3 = 9.0 x 10 -12 s = 1.3 x 10 -4 mol/L For uni-bivalent type of electrolytes

7. Mg(OH) 2 (s) ⇌ Mg 2+ (aq) + 2 OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 8.9 x 10 -12 If s is the solubility of Mg(OH) 2 , then: [Mg 2+ ] = s mol/L and [OH - ] = 2 s mol/L, K sp = ( s )(2 s ) 2 = 4 s 3 = 8.9 x 10 -12 s = 1.3 x 10 -4 mol/L Solubility and Solubility Products For Bi-univalent type of electrolytes

8. Ag 3 PO 4 (s) ⇌ 3Ag + (aq) + PO 4 3- (aq) K sp = [Ag + ] 3 [PO 4 3- ] = 1.8 x 10 -18 If s is the solubility of Ag 3 PO 4 , then: K sp = (3 s ) 3 ( s ) = 27 s 4 = 1.8 x 10 -18 s = 1.6 x 10 -5 mol/L Solubility and Solubility Products For uni-trivalent type of electrolytes

9. Cr(OH) 3 (s) ⇌ Cr 3+ (aq) + 3 OH - (aq) K sp = [Cr 3+ ][OH - ] 3 = 6.7 x 10 -31 If the solubility is s mol/L, then: K sp = [Cr 3+ ][OH - ] 3 = (s)(3s) 3 = 27s 4 = 6.7 x 10 -31 s = 1.3 x 10 -8 mol/L Solubility and Solubility Products For tri-univalent type of electrolytes

10. Ca 3 (PO 4 ) 2 ( s ) ⇌ 3Ca 2+ (aq) + 2PO 4 3- (aq) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 = 1.3 x 10 -32 If the solubility is s mol/L, then: [Ca 2+ ] = 3 s , and [PO 4 3- ] = 2 s K sp = (3s) 3 (2s) 2 = 108 s 5 = 1.3 x 10 -32 s = 1.6 x 10 -7 mol/L Solubility and Solubility Products For Bi-trivalent type of electrolytes

11. General expression: A x B y (s) ⇄ x A y + (aq) + y B x - (aq) x s y s Solubility product, K sp = [A y + ] x [ B x - ] y = [ x s] x [ y s] y = x x y y [s] x+y Solubility and Solubility Products s = K sp x x y y 1/ x+y

12. Solubility and Ksp Three important definitions: Solubility : Quantity of a substance that dissolves to form a saturated solution Molar solubility : The number of moles of the solute that dissolves to form a liter of saturated solution Ksp (solubility product) : The equilibrium constant for the equilibrium between an ionic solid and its saturated solution

13. Some Values For Solubility Product Constants (K sp ) At 25 o C

14. Problem Write the expressions of K sp of: a) AgCl b) PbI 2 c) Ca 3 (PO 4 ) 2 d) Cr(OH) 3

15. Problem If a saturated solution of BaSO 4 is prepared by dissolving solid BaSO 4 in water, and [Ba 2+ ] = 1.05 x 10 -5 mol  L -1 , what is the K sp for BaSO 4 ?

16. The Common-ion Effect The suppression of the dissociation of weak electrolyte in the presence of strong electrolyte having an ion common with that of the former is known as “common-ion effect” Consider the following solubility equilibrium: AgCl (s) ⇌ Ag + ( aq ) + Cl - ( aq ) ; K sp = 1.6 x 10 -10 ; The solubility of AgCl is 1.3 x 10 -5 mol/L at 25 o C. If NaCl is added, equilibrium shifts left due to increase in [ Cl - ] and some AgCl will precipitate out.

17. Common Ion Effect on Solubility The presence of a common ion decreases the solubility of a slightly soluble ionic compound. PbCrO 4 ( s ) Pb 2 + ( aq ) + CrO 4 2 - ( aq ) Add Na 2 CrO 4 (a very soluble salt; a strong electrolyte) Result : Equilibrium shifts to the left (LeChâtelier’s principle) The effect of a common ion on solubility PbCrO 4 ( s ) Pb 2 + ( aq ) + CrO 4 2 - ( aq ) PbCrO 4 ( s ) Pb 2 + ( aq ) + CrO 4 2 - ( aq ) CrO 4 2 - added

18. Aqua Regia Aqua regia is a mixture of concentrated HCl and concentrated HNO 3 . Either one of these acids alone will not affect Au. The mixture will dissolve gold--one of the most inactive metals. What does aqua regia mean? Royal Water

19. The solubility product constant, K sp , represents equilibrium between a slightly soluble ionic compound and its ions in a saturated aqueous solution. The common ion effect is responsible for the reduction in solubility of a slightly soluble ionic compound. The solubilities of some slightly soluble compounds depends strongly on pH. Because some salts will not dissolve well in pure water, but will dissolve in an acid or a base. Summary