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Chapter 18

Chapter 18. Equilibria contd. K sp. Overview. Common Ion Effect Buffers Solubility. Common Ion Effect. Example: Na Ac and HAc NaAc → Na +1 + Ac -1 100% ionized

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Chapter 18

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  1. Chapter 18 Equilibria contd. Ksp

  2. Overview • Common Ion Effect • Buffers • Solubility

  3. Common Ion Effect • Example: Na Ac and HAc • NaAc → Na +1 + Ac -1 100% ionized • HAc ↔ H +1 + Ac -1 Le Chatelier predicts, if [Ac -1 ] incr, shift ← H +1 is lowered or pH increases ( ex. p. 812)IONIZATION IS SUPRESSED BY COMMON ION

  4. BUFFERS • Defined:A buffer solution is a special case of the common ion effect. The function of a buffer is to resist changes in the pH of a solution. • Components: wa / salt of (cb) or wb/ caex. HAc + H2O ↔ H3O+ + Ac- wa and NaAc 100% →Na + +Ac-So, [Ac- ] is large

  5. Buffer, sample problem Problem: What is the pH of a buffer that has [HAc] = 0.700 M and [Ac-] = 0.600 M? HAc + H2O ↔ Ac- + H3O+ Ka = 1.8 x 10-5 [HAc] ↔ [Ac- ] + [H3O+] At equilib 0.700 - x 0.600 + x x Since 100K < 0.700 and 0.600, we have [H3O+] = 2.1 x 10-5 and pH = 4.68

  6. Buffer Action • Write reactions-acid added/ base addedKa =[H3O+ ] [Ac-] [HAc] Solve, [H3O+ ] = Ka [HAc] [Ac-]this ratio changes only slightly in a buffered solution

  7. Henderson-Hasselbalch Equation • Convenient for finding correct wa and salt to buffer to a given pH • [H3O+ ] = Ka [HX] [X-] taking – log of both sides, pH = pKa + log [ X- / HX] Look up pKa - see Appx H or table 17.3- find correct acid for given pH and use equimolar concs of acid and salt ..ie, set [X- ]= [HX]

  8. Sample Problem Problem: If you use HAc/Ac- to prepare a buffer of pH= 4.3, what should be the ratio of [HAc]/[Ac-] ?Solution: first, look up pKa = 4.74 Use H-H eqn: pH = pKa + log [ X- / HX] Plug in, 4.3 = 4.74 + log [ Ac- / HAc]Solve, log [ Ac- / HAc] = 4.3 – 4.74 = - .44 [ Ac- / HAc] = 10 -.44 = 0.36 and [HAc]/[Ac-] = 1/0.36 = 2.8 ans.

  9. Review Terms • Unstaurated • Saturated • Solubility • “insoluble” vs “soluble”

  10. Solubility Product, Ksp Saturated BaSO4 BaSO4 ↔ Ba+2 + SO4-2 Ksp= [Ba+2 ] [SO4-2 ] = 1.1 x 10 -10 @ 250C see p. 834 and Appx JKsp= [x][x] = x2 = 1.1 x 10 -10 x = 1.0 x 10-5M solubility

  11. “Solubility”&Solubility Product Constant,Ksp • Compare: AgCl to Ag2CrO4Ksp= [Ag+1 ] [Cl-1 ] = [x][x]= x 2 = 1.8 x 10 -10 solubility, x= 1.3 x 10 -5 M Ksp= [Ag+1 ]2 [CrO4-2 ] = [2x]2 [x] = 4x3 = 9 x 10 -12 solubility,x = 1.4 x 10 -4 M Ksp & solubility compare ONLY when +/- ratios SAME

  12. Recall, common ion Lowers solubility- Illustration:What’s solubility of AgCl in 1.0 L of 0.55 M NaCl?AgCl ↔ Ag+1 + Cl-1 Initial 0 0.55Change +x +xEq x 0.55 + x(can assume x is small: 100K = 1.8 x 10 -8 < 0.55 )Ksp= [x] [0.55 +x] = [x] [0.55]= 1.8 x 10 -10x = 3.3 x 10 -10 M note, in pure water x = 1.3 x 10 -5 M

  13. Basic Anions Increase Salt Solubility Ex. PbS ↔ Pb +2 + S -2 But S -2 is a “basic” anion- i.e.- it can hydrolyze: S -2 + H2O ↔ HS -1 + OH -1 PbS dissolves as S -2 is used (Le Chatelier prediction) Note, S -2 is the cb of the wa, HS -1 A salt with an anion that is the cb of a wa, will dissolve more than its Ksp predicts. These salts can also dissolve in sa, since:2H3O+ + S -2 ↔ H2S + 2H2O

  14. Predicting PPT • IP > Ksp ppt forms • IP = Ksp solution is just saturated • IP < Ksp ppt can NOT form

  15. Dilution Effects on PPT Illustrative ComputationMix 0.100 L 0.0015M MgCl2 and 0.200 L 0.025M NaFGiven Ksp MgF2 = 3.7 x 10 -8 Approach: 1st, compute ion concs; next compute IP; last, compare IP to given Ksp …….try this at home!

  16. Completeness of PPT standard: 0.1% of target ion remains in solution ( ex, Mg+2 as “target” for F-1)Completeness favored by: • very small Ksp value • High initial conc of target ion • Conc of common ion >> conc target ionEx.: MgF2 if Mg+2 is target ( and F -1 is common ion)

  17. Selective PPT AgClKsp= [Ag+1 ] [Cl-1 ] = 1.8 x 10 -10 AgIKsp= [Ag+1 ] [I-1] = 8.5 x 10 -17 It is possible to pptAgI w/o pptAgCl and thereby separate the two anions? The greater the difference by orders of 10, of the Ksp values, more likely can selectively ppt

  18. Example, selective ppt A solution contains 0.020 M Ag+ and 0.02M Pb2+. Add CrO42- to precipitate Ag2CrO4 and PbCrO4. Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10-12 Ksp for PbCrO4 = 1.8 x 10-14 Solution Calculate [CrO42-] required by each ion. [CrO42-] to ppt. PbCrO4 = Ksp/ [Pb2+] [CrO42-] = 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M [CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2 [CrO42-] = 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M PbCrO4ppts first !What’s the [Pb2+], when Ag2CrO4 starts to ppt? Is PbCrO4pptcomplete?

  19. pH and Solubility • Case 1) CaF2 and SA CaF2 ↔ Ca+2 + 2F-1 H3O+1 + F-1 ↔ HF + H2O Adding H3O+1 (from a SA) shifts both equilibria → therefore, CaF2 has increased solubility in acid solution; is pH effected • Case 2) AgCl and SA H3O+1 + Cl-1 → ???

  20. End Chapter 18 You read: • Section 18.3 Titrations • Section 18.6 & 18.7 Complex Ions

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