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Gravimetric analysis

Gravimetric analysis. Chemistry 321, Summer 2014. Analytical scheme:. A ( aq ) + R ( aq ) analyte reagent . AR (s) neutral complex precipitate ( ppt ) .

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Gravimetric analysis

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  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 purifyppt, before drying and weighing. Use the mass of ppt and rxnstoichiometry to determine the mass of A in sample.

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

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

  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 solubilityproduct, Ksp Ksp where Ksp = [Ag+] [Cl–] Why is [AgCl (s)] left out?

  7. Consider the actual equation from which Ksp is derived: K°spis the thermodynamic equilibrium constant where ai = the activity of species i = fi [i] and fi 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°spare tabulated at zero ionic strength (μ) where fi = 1 so K°sp = Ksp 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 AxBy Ksp AxBy(s) x Ay+(aq) + y Bx–(aq) Ksp = [Ay+]x[Bx–]y Precipitation occurs from an aqueous solution when [Ay+]x[Bx–]y≥ Ksp

  10. Solubility Solid AgCl is added to pure H2O and the resulting solution is allowed to come to equilibrium Ksp= [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. Ksp= s × s = s2 = 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 AgNO3 and recalculate s

  13. Ksp = (0.001+ s) (s) = 1 × 10–10 (recall that Ksp 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 H2O. 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?

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