1. Gravimetric analysis Chemistry 321, Summer 2014

2. Analytical scheme: A (aq) + R (aq) analyte reagent AR (s) neutral complex precipitate (ppt) Goal: Isolate and purify ppt, before drying and weighing. Use the mass of ppt and rxn stoichiometry to determine the mass of A in sample.

3. Analytical scheme: A (aq) + R (aq)AR (s) KfKf = formation constant K f needs to be large for rxn to go to completion

4. Examples nickel ion (analyte) dimethylglyoxime (DMG) (reagent) Ni(DMG) 2 (solid precipitate) silver ion (reagent) chloride ion (analyte) silver chloride (solid precipitate) Note: you must control solution pH!

5. Can get good accuracy and precision by gravimetric analysis Analytic balance precision is 0.1 mg in 100 g of sample 0.1 × 10 –3 /100 = 10 –6 uncertainty

6. Note that the formation constant is the reciprocal of the solubility product, K sp K sp where K sp = [Ag + ] [Cl – ] Why is [AgCl (s)] left out?

7. Consider the actual equation from which K sp is derived: K° sp is the thermodynamic equilibrium constant where a i = the activity of species i = f i [i] and f i is the activity coefficient for i and [i] is measured in molar For a solid or a pure liquid at standard state, a = 1

8. The K° sp are tabulated at zero ionic strength (μ) where f i = 1 so K° sp = K sp Conclusion: Use tabulated K° sp to perform calculations…generally it’s okay to do so even if f is not exactly 1; located in Table C3, pp. 806–807.

9. General Expression for Salt A x B y A x B y (s) K sp x A y+ (aq) + y B x– (aq) K sp = [A y+ ] x [B x– ] y Precipitation occurs from an aqueous solution when [A y+ ] x [B x– ] y ≥ K sp

10. Solubility Solid AgCl is added to pure H 2 O and the resulting solution is allowed to come to equilibrium K sp = [Ag + ] [Cl – ] = 1 × 10 –10 (from table) Set up an ICE table (recall general chemistry equilibrium) for this system and define s = solubility of AgCl (M)

11. AgCl (s)Ag+ (aq)Cl– (aq) initial (M)—00 change (M)—ss equil (M)—ss K sp = s × s = s 2 = 1 × 10 –10 so s = [Ag + ] = [Cl – ] = 1 × 10 –5 M

12. Common ion effect A ppt is more soluble in pure water than in a solution containing one or more of the ions in the ppt. Example: To the saturated AgCl solution from the previous slide, add 0.001 M AgNO 3 and recalculate s AgCl (s)Ag+ (aq)Cl– (aq) initial (M)—0.0010 change (M)—ss equil (M)—0.001 + ss

13. K sp = (0.001+ s) (s) = 1 × 10 –10 (recall that K sp is a constant!) To avoid using the quadratic formula to solve for s, make a simplifying assumption: s << 0.001 M (we shall have to check this assumption) so 0.001 + s ≈ 0.001, and (0.001) (s) ≈ 1 × 10 –10 so s = [Cl – ] = 1 × 10 –7 M check assumption: 1 × 10 –7 M << 0.001 M

14. Adding Ag + results in a significant reduction in Cl – solubility In 1 mM Ag +, the chloride ion is 100 times less soluble than in pure water Practical consequence: By adding excess Ag +, Cl – is more effectively precipitated, which helps reduce measurement error

15. Challenge problem (next Wednesday) The solubility of Cl – is 1 × 10 –5 M with AgCl in otherwise pure H 2 O. How can gravimetry be used to determine [Cl – ] in aqueous samples if [Cl – ] is thought to be ≈ 10 –6 M? Hint: The goal is to get 99.9% of the chloride ion to precipitate (in other words, 0.1% Cl – left) Apply the common ion effect – what would you add to the sample to meet the goal?