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Ch15 z5e aq. equil

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  • 1. Chapter 15 Applying equilibrium pp
  • 2. 15.1 The Common Ion Effect pp When the salt with the anion of a weak acid is added to that acid, It reverses the dissociation of the acid. Lowers the % dissociation of the acid. HAc H 1+ + Ac 1- (add NaAc) Same for salts with cation of weak base . H 2 O + NH 3 NH 4 1+ + OH 1- (add NH 4 Cl) The calculations are the same as last chapter.
  • 3. 15.2 Buffered solutions pp A solution that resists a change in pH. Consists of either a weak acid and its salt or a weak base and its salt . We can make a buffer of any pH by varying the concentrations of these solutions. Same calculations as before. Calculate pH of a solution of .50 M HAc & .25 M NaAc (Ka = 1.8 x 10 -5 ). Steps follow.
  • 4. pH of .50 M HAc & .25 M NaAc pp Write major species: HAc, Na 1+ , Ac 1- , H 2 O R HAc H 1+ + Ac 1- I 0.50 0 0.25 C - x + x + x E 0.50 - x x 0.25 + x Assume x <<, since K a = 1.8 x 10 -5 (x)(.25)/(.50) = 1.8 x 10 -5 [H 1+ ] = ??? [H 1+ ] = 3.6 x 10 -5 pH = ??? . . . pH = 4.44
  • 5. Adding a strong acid or base pp Do the stoichiometry first (use moles ). Think about the chemistry, write major species & check to see an =m reaction will occur among them. A strong base will grab protons from the weak acid reducing [HA] 0 A strong acid will add its proton to the anion of the salt reducing [A - ] 0 Then do the =m problem in molarity .
  • 6. Another way to think of it pp Write major species & look for a strong something & a weak something. The stoich will be between them (SA + WB or SB + WA) in moles . E.g., If major species are HAc, Ac 1- , H 1+ (from HCl) & H 2 O, the stoich is . . . H 1+ + Ac 1- HAc SA WB WA Then do the =m problem in molarity .
  • 7. Adding a strong acid or base to buffered soln pp What is pH of 1.0 L of 0.50 M HAc + 0.25 M NaAc after add 0.010 mol solid NaOH? Write major species . . . HAc Na 1+ Ac 1- OH 1- H 2 O Look for a stoich reaction in moles . . . HAc + OH 1- HOH + Ac 1- i 0.50 0.01 0.25 mol f 0.49 0 0.26
  • 8. Adding a strong acid or base pp HAc + OH 1- HOH + Ac 1- i 0.50 0.01 0.25 mol f 0.49 0 0.26 Now change back to M (had 1 L) & do =m R HAc H 1+ + Ac 1- I 0.49 0 0.26 C - x + x + x E 0.49 - x x 0.26 + x
  • 9. Adding a strong acid or base pp R HAc H 1+ + Ac 1- I 0.49 0 0.26 C - x + x + x E 0.49 - x x 0.26 + x Assuming x << than 0.49 or 0.26 (x)(0.26)/(0.49) = 1.8 x 10 -5 X = [H 1+ ] = ??? 3.4 x 10 -5 pH = ??? 4.47 (compare to 4.44 before adding OH 1- )
  • 10. What if there was no buffer? pp What is pH of 1.0 L of 0.50 M HAc after adding 0.010 mol of solid NaOH? (pH of HAc before adding OH 1- is 2.52) Write major species . . . HAc Na 1+ OH 1- H 2 O Look for a stoich reaction in moles . . . HAc + OH 1- HOH + Ac 1- i 0.50 0.01 0 mol f 0.49 0 0.01
  • 11. What if there was no buffer? pp HAc + OH 1- HOH + Ac 1- i 0.50 0.01 0 mol f 0.49 0 0.01 Now to M & do =m; K a = 1.8 x 10 -5 R HAc H 1+ + Ac 1- I 0.49 0 0.01 C - x + x + x E 0.49 - x x 0.01 + x pH = ??? pH = 3.05, compare to 2.52 for HAc alone.
  • 12. General equation Ka = [H + ] [A - ] [HA] so [H + ] = Ka [HA] [A - ] The [H + ] depends on the ratio [HA]/[A - ] taking the negative log of both sides pH = -log(Ka [HA]/[A - ]) pH = -log(Ka) + -log([HA]/[A - ]) pH = pKa + log([A - ]/[HA])
  • 13. This is called the Henderson-Hasselbach equation pH = pK a + log([A - ]/[HA]) pH = pK a + log([base]/[acid]) Only use this (instead of RICE) when there is a BUFFER! It’s a nice short cut for buffers, but students tend to use it incorrectly for other problems, so be careful.
  • 14. Sample Problem pp pH = pK a + log([base]/[acid]) Use H/H to calculate the pH of 0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (K a = 1.4 x 10 -4 ) . . . pK a = ??? pK a = 3.85 pH = 3.85 + log (0.25/0.75) = ??? pH = 3.37 Can also do by the RICE method.
  • 15. Another Sample Problem pp pH = pK a + log([base]/[acid]) Calculate the pH of 0.25 M NH 3 and 0.40 M NH 4 Cl (K b = 1.8 x 10 -5 ) . . . Whoa! We have K b . Are we stuck? No! K a x K b = K w = 1.0 x 10 -14 Your answer is . . . pH = 9.05 Calculation . . . pH = (14 - pK b ) + log(.25/.40) = 9.05 Can get the same using RICE, but will get pOH and must subtract from 14 to get pH.
  • 16. Hints for doing the online HW pp Most of the problems will have a part to them that you can solve using the H-H equation . If there is a neutralization reaction then do the stoichiometry first to determine how the mmoles of HA and A 1- have changed. Then, convert to [ ] and use the H-H equation . Remember that if you have equal equilibrium amounts of HA and A 1- then pk a = pH .
  • 17. Hints for doing the online HW pp Acids & Bases dissociate stepwise , so even if you don’t know the acid (e.g, benzoic acid) you know that it loses only one hydrogen (unless you’re told it is diprotic, triprotic, etc.) The term “ ionization constant ,” as used in the online HW, is a generic term for k a or k b
  • 18. Let’s Prove They’re buffers pp What is the pH if .10 mol of gaseous HCl is added to 1.0 L of 0.25 M NH 3 & 0.40 M NH 4 Cl (K b = 1.8 x 10 -5 )? (Why gaseous? Why 1.0 L?) Gaseous or solid means total volume doesn’t change. 1.0 L means mols = molarity Be careful! Convert mols back to molarity since volume might change!! See summary, p. 692
  • 19. Let’s Prove They’re buffers pp What is the pH if .10 mol of gaseous HCl is added to 1.0 L of 0.25 M NH 3 & 0.40 M NH 4 Cl (K b = 1.8 x 10 -5 )? Steps follow. Do the stoich to get . . . H 1+ + NH 3 NH 4 1+ i .10 .25 .40 f 0 .15 .50
  • 20. Let’s Prove They’re buffers pp pH if .10 mol HCl (g) added to 1.0 L of 0.25 M NH 3 & 0.40 M NH 4 Cl K b = 1.8 x 10 -5 After stoich, 0.15 M NH 3 & 0.50 M NH 4 1+ Do the Equilibrium to get . . . R NH 3 + H 2 O NH 4 1+ + OH 1- I .15 .50 0 C - x +x +x E .15 - x .50 + x x Etc. or Use Henderson-Hasselbach! . . .
  • 21. Let’s Prove They’re buffers pp pH if .10 mol HCl (g) added to 1.0 L of 0.25 M NH 3 & 0.40 M NH 4 Cl K b = 1.8 x 10 -5 Using H-H pH = -log(10 -14 /1.8 x 10 -5 ) + log(.15/.50) pH = 8.73 (Compare to original of 9.05)
  • 22. Let’s Prove They’re buffers pp What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of 0.75 M lactic acid HC 3 H 5 O 3 ) & 0.25 M sodium lactate (K a = 1.4 x 10 -4 ) Do Stoich, then H-H equation to get ??? pH = 3.48, compare to original of 3.38. Be careful! Convert mols back to molarity since volume might change!!
  • 23. 15.3 Buffer capacity The pH of a buffered solution is determined by the ratio [A - ]/[HA]. As long as this ratio doesn’t change much the pH won’t change much. The more concentrated these two are the more H + and OH - the solution will be able to absorb. Larger concentrations means bigger buffer capacity.
  • 24. Buffer Capacity - Use “RIF” & H-H method for these Calculate pH change when 0.010 mol HCl (g) is added to 1.0L solution s of HAc & NaAc . . . (Ka = 1.8x10 -5 & pH without HCl = 4.74) Write the reaction (SA + WB), which is . . . H 1+ + Ac 1- HAc (HAc not always on reactant side) 5.00 M HAc & 5.00 M NaAc. pH = ??? pH still 4.74 (insignificant change) 0.050 M HAc & 0.050 M NaAc. pH = ??? pH = 4.56 so change is only -0.18
  • 25. Buffer capacity The best buffers have a ratio [A - ]/[HA] = 1 Memorize this rule of thumb! This is most resistant to change True when [A - ] = [HA] Make pH = pKa (since log1 = 0)
  • 26. Buffer capacity Which acid (& its sodium salt) below will give the best buffer at pH = 4.30? Chloracetic acid (Ka = 1.35 x 10 -3 ) Propanoic acid (Ka = 1.3 x 10 -5 ) Benzoic acid (Ka = 6.4 x 10 -5 ) Hypochlorous acid (Ka = 3.5 x 10 -8 ) . Hint: don’t find actual [ ]s of HA and A - ; just their ratio. Answer ??? . . . Benzoic Acid (pK a = 4.19, closest to 4.30)
  • 27. 15.4 Titrations & pH Curves Millimole (mmol) = 1/1000 mol Molarity = mmol/mL = mol/L Makes calculations easier because we will rarely add Liters of solution. Adding a solution of known concentration until the substance being tested is consumed. This is called the equivalence point . Graph of pH vs. mL is a titration curve.
  • 28. Strong acid with Strong Base pp Do the stoichiometry . There is no equilibrium . They both dissociate completely. Always write the major species that exist before any reaction occurs!! Let’s titrate 50.0 mL of 0.200 M HNO 3 with 0.100 M NaOH with the following volumes (watch volume changes) . . . Analyze the pH after adding total NaOH to 0.0, 10.0, 20.0, 100.0, 150.0 ml . . .
  • 29. Strong acid with Strong Base pp Titration of 50.0 mL of 0.200 M HNO 3 with 0.00 of 0.100 M NaOH What’s pH? Major species? H 1+ NO 3 1- H 2 O . . . Stoich reaction? . . . Nope, so pH = -log [H 1+ ] = -log[.2] = .699
  • 30. Strong acid with Strong Base pp Titration of 50.0 mL of 0.200 M HNO 3 with 10 .0 mL of 0.100 M NaOH What’s pH? Major species? H 1+ NO 3 1- H 2 O Na 1+ OH 1- . . . Reaction? R H 1+ + + OH 1- H 2 O , do “RIF” I 10 mmol 1 mmol (note: mmol, not molarity) F 9 0 [H 1+ ] = 9 mmol/how much solution? [H 1+ ] = 9 mmol/ 60 mL = 0.15 M , pH = ?? pH = 0.82
  • 31. Strong acid with Strong Base pp Titration of 50.0 mL of 0.200 M HNO 3 with 20 .0 of 0.100 M NaOH What’s pH? Major species = H 1+ NO 3 1- H 2 O Na 1+ OH 1- R H 1+ + OH 1- H 2 O , do “RIF” I 10 mmol 2.0 mmol F 8 0 [H 1+ ] = 8 mmol/ 70 mL = 0.11 M pH = 0.942 Note: the volume is different from when we previously added only 10.0 mL NaOH
  • 32. Strong acid with Strong Base pp Titration of 50.0 mL of 0.200 M HNO 3 with 100 .0 of 0.100 M NaOH What’s pH? Major species = H 1+ NO 3 1- H 2 O Na 1+ OH 1- R H 1+ + + OH 1- H 2 O , do “RIF” I 10 mmol 10 mmol F 0 0 Major species now = NO 3 1- H 2 O Na 1+ Only =m is H 2 O H 1+ + + OH 1- pH = 7.00
  • 33. Strong acid with Strong Base pp Titration of 50.0 mL of 0.200 M HNO 3 with 150 .0 of 0.100 M NaOH What’s pH? Major species = H 1+ NO 3 1- H 2 O Na 1+ OH 1- R H 1+ + + OH 1- H 2 O , do “RIF” I 10 mmol 15 mmol F 0 5 Major species now = NO 3 1- H 2 O Na 1+ OH 1- [ OH 1- ] = 5 mmol/200ml = 0.025 M , pH = ? pH = 12.40 (remember to subtract p OH from 14.00!!)
  • 34. Strong acid with strong Base pp Equivalence at pH 7 pH mL of Base added 7
  • 35. Weak acid with Strong base pp There is an equilibrium. Do stoichiometry first . Then do equilibrium. Titrate 50.0 mL of 0.10 M HAc (Ka = 1.8 x 10 -5 ) with 0.10 M NaOH adding to a total addition volume of 0.0, 10.0, 25.0, 50.0 and 60.0 ml sequentially.
  • 36. Weak acid with Strong base pp Titrate 50.0 mL of 0.10 M HAc with 0.00 mL of 0.10 M NaOH. Major species? HAc H 2 O, Stoich??? No. So go to “RICE” R HAc H 1+ + Ac 1- I .10 M 0 0 C - x +x +x E .10 - x x x pH = ??? pH = 2.87
  • 37. Weak acid with Strong base pp Titrate 50.0 mL of 0.10 M HAc with 10.00 mL of 0.10 M NaOH. Major species? HAc Na 1+ OH 1- H 2 O, Stoich??? Yes. Strong base & weak acid . “RIF” R HAc + OH 1- HOH + Ac 1- I 5 mmol 1 mmol 0 F 4 0 1 What are major species now?
  • 38. Weak acid with Strong base pp HAc Na 1+ Ac 1- H 2 O, Stoich??? No . No Strong base. RICE or H-H eq R HAc + H 1+ + Ac 1- I 4/ 60 0 1/ 60 C - x + x + x E (4/60) - x + x (1/60) + x Simplifying: (x)(1/60)/(4/60) = 1.8 x 10 -5 pH = ??? pH = 4.14 H/H: pH = pKa + log([Ac - ]/[HAc]) = 4.14
  • 39. Weak acid with Strong base pp Titrate 50.0 mL of 0.10 M HAc with 25.00 ml of 0.10 M NaOH. Major species? HAc Na 1+ OH 1- H 2 O, Stoich??? Yes. Strong base & weak acid. “RIF” R HAc + OH 1- HOH + Ac 1- I 5 mmol 2.5 mmol 0 F 2.5 0 2.5 What are major species now? . . .
  • 40. Weak acid with Strong base pp HAc Na 1+ Ac 1- H 2 O, Stoich??? No. No Strong base. RICE or H-H eq R HAc + H 1+ + Ac 1- I 2.5/ 75 0 2.5/ 75 C - x + x +x Can use RICE or H-H or . . . Since [HAc] = [Ac 1- ], pH = pKA pH = ??? pH = 4.74
  • 41. Weak acid with Strong base pp Titrate 50.0 mL of 0.10 M HAc with 50.00 ml of 0.10 M NaOH. Major species? HAc Na 1+ OH 1- H 2 O, Stoich??? Yes . Strong base & weak acid. “RIF” R HAc + OH 1- HOH + Ac 1- I 5 mmol 5 mmol 0 F 0 0 5 What are major species now?
  • 42. Weak acid with Strong base pp Na 1+ Ac 1- H 2 O, Stoich??? No . No Strong base. RICE or H-H eq What is the =m reaction & RICE? R Ac 1- + H 2 O HAc + OH 1- I 5/ 100 0 0 C - x + x +x E .05 - x M x x x 2 /(.05) = k b = 10 -14 /1.8 x 10 -5 pH = ??? pH = 8.72 (14 - pOH)
  • 43. Weak acid with Strong base pp Titrate 50.0 mL of 0.10 M HAc with 60.00 mL of 0.10 M NaOH. Major species? HAc Na 1+ OH 1- H 2 O, Stoich??? Yes. Strong base & weak acid. “RIF” R HAc + OH 1- HOH + Ac 1- I 5 mmol 6 mmol 0 F 0 1 5 What are major species now?
  • 44. Weak acid with Strong base pp Na 1+ Ac 1- H 2 O, OH 1- Have 2 bases, but OH 1- is stronger. Amount OH 1- from Ac 1- & H 2 O is small (negligible compared to that from NaOH). [OH 1- ] = 1.0 mmol/110 ml pH = ??? pH = 11.96 (14 - pOH)
  • 45. Diprotic Acid pp Calculate [H + ] after 75.0 ml of 0.20 M NaOH has been added to 100.0 ml of 0.10 M H 2 A Ka 1 = 1.5 x 10 -4 Ka 2 = 8.0 x 10 -7 Steps . . . Treat each H 1+ as being separately reacted with OH 1-
  • 46. Diprotic Acid pp [H + ] with 75.0 ml of 0.20 M NaOH + 100.0 ml of 0.10 M H 2 A (Ka 1 = 1.5 x 10 -4 ; Ka 2 = 8.0 x 10 -7 ). R H 2 A + OH 1- HA 1- + HOH I 10 15 F 0 5 10 R HA 1- + OH 1- A 2- + HOH I 10 5 0 F 5 0 5 R HA 1- H 1+ + A 2- ICE 5-x x 5+x . . .
  • 47. Diprotic Acid pp [H + ] with 75.0 ml of 0.20 M NaOH + 100.0 ml of 0.10 M H 2 A (Ka 1 = 1.5 x 10 -4 ; Ka 2 = 8.0 x 10 -7 ). R HA 1- H 1+ + A 2- ICE 5 x 5 Since [HA 1- ] = [A 2- ], pH = pKa 2 = 6.10 You will have a problem like this on the test!
  • 48. Titration Curves
  • 49. Strong acid with strong Base Equivalence at pH 7 pH mL of Base added 7
  • 50. Titration Curve: Strong Acid & Strong Base Over what pH range does the titrated solution suddenly change from high [H 3 O 1+ ] to high [OH 1- ] About 2.5 to 12 Look at page 756 to find the best indicator . . . Bromothymol Blue (its midpoint is close to the equivalent point)
  • 51. pH mL of Base added >7 Weak acid with strong Base Equivalence at pH >7
  • 52. Titration Curve for Weak Acid & Strong Base Over what pH range does the titrated solution suddenly change from high [H 3 O 1+ ] to high [OH 1- ]? About 6.5 to 11 What’s pH when 0.067 mol NaOH added? pH = 5 What’s [H 3 O 1+ ]? [H 3 O 1+ ] = 1 x 10 -5 M Best Indicator (p. 756 & next slide)? o -Cresolphthalein or Phenolphthalein.
  • 53. Useful pH ranges for several common indicators. p . 715
  • 54. Equivalence point (endpoint) Point at which equal amounts of H 3 O + and OH - have been added. Determined by… indicator color change Titration dramatic change in pH pK a is 1/2-way point
  • 55. pH mL of Base added 7 Strong base with strong acid Equivalence at pH 7
  • 56. pH mL of Base added <7 Weak base with strong acid Equivalence at pH <7
  • 57. Summary - titrating acids pp Strong acid and base just stoichiometry. Weak acids: determine K a , use for 0 mL base added Weak acid before equivalence point Stoichiometry first Then Henderson-Hasselbach or RICE No weak acid at equivalence point: use K b and get pH by 14 - pOH Have weak base after equivalence, but use leftover strong base since it predominates.
  • 58. Summary - titrating bases pp Determine K b , use for 0 mL acid added. Calc. pOH, then subtract from 14 for pH Weak base before equivalence point. Stoichiometry first Then Henderson-Hasselbach Subtract pOH from 14 for pH No weak base at equivalence point; so use K a . Have weak acid after equivalence, but use leftover strong acid since it predominates.
  • 59. 15.5 Acid/Base Indicators Weak acids that change color when they become bases. weak acid written HIn HIn H + + In - clear red Equilibrium is controlled by pH End point - when the indicator changes color. Transition interval - pH range over which an indicator changes color. See next slide.
  • 60. Figure 15.8 p. 756: The Useful pH Ranges for Several Common Indicators
  • 61. Color Ranges of Various Indicators Used in Titrations Why are different indicators used for SA-SB, SA-WB, or WA-SB titrations? Each type of titration has a different end point. Is it better to use an indicator with a color change over a wide or narrow pH range. Why? Narrow pH range since gives better approximation of the end point.
  • 62. pH Scale Relationships If [H 3 O 1+ ] of “x” is 10 x greater than another with pH 3, what is pH of “x”? pH = 2 If [OH 1- ] of “y” is 1000 x greater than pure water’s, what is pH of “y”? pH = 10 Is hand soap more or less basic than baking soda? More basic.
  • 63. Figure 15.9 p. 757 The pH Curve for the Titration of 100.0 mL of 0.10 M HCI with 0.10 M NaOH Which is the better indicator? Neither, since their ranges are not close to the equivalence (end) point.
  • 64. Figure 15.10 p. 757 The pH Curve for the Titration of 50 mL of 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH Choose the best indicator between the two . . . Phenolphthalein
  • 65. Indicators pp Moving towards equilibrium causes a gradual color change . It is noticeable when the ratio of [In - ]/[HI] or [HI]/[In - ] is 1:10 ( 1/10 ) The Indicator is a weak acid , so it has a K a . The pH the indicator changes at is when [In - ]/[HI] = 1/10 . Using H-H . . . That pH = pKa +log([In - ]/[ HI ]) = pKa +log(1/10) So, pH at which the indicator changes = pKa - 1 on the way up the pH scale (i.e., titrating an acid using a base).
  • 66. Indicators pp pH=pKa + log([ HI ]/[In - ]) = pKa + log(10) pH = pK a + 1 on the way down (i.e., titrating a base by adding an acid). Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating an acid with base . Choose the indicator with a pKa 1 greater than the pH at equivalence point if you are titrating a base with acid .
  • 67. 15.6 Solubility Equilibria Will it all dissolve, and if not, how much? Note on this week’s online HW: One or more questions contains more information than you need. You have to discriminate between what information is relevant. Strategy: Given all the information figure out your plan and discard information not needed for that plan. Have fun!
  • 68. All dissolving is an equilibrium . If there is not much solid it will all dissolve. As more solid is added the solution will become saturated. Solid dissolved Solid will precipitate as fast as it dissolves. So, it is a dynamic equilibrium Remember : pure liquids and pure solids are never included in an equilibrium expression. Solubility Equilibria
  • 69. General equation M + stands for the cation (usually metal). Nm - stands for the anion (a nonmetal). M a Nm b (s) a M + (aq) + b Nm - (aq) K = [M + ] a [Nm - ] b /[M a Nm b ] But the concentration of a solid doesn’t change (so drops out of the =m expression). K sp = [M + ] a [Nm - ] b This is called the solubility product for each compound.
  • 70. Watch out Solubility is not the same as solubility product . Solubility product is an equilibrium constant . It does NOT change ( except with temperature). Solub ility is an equilibrium position for how much can dissolve (so can shift). A common ion can change solubility .
  • 71. Calculating K sp The solubility of iron(II) oxalate FeC 2 O 4 is 65.9 mg /L The solubility of Li 2 CO 3 is 5.48 g /L We would use the solubility to then calculate K sp for each of them.
  • 72. Calculating Solubility pp The solubility is determined by the equilibrium. It is an equilibrium problem (RICE). Calculate the K sp of Bi 2 S 3 , with a solubility of 1.0 x 10 -15 mol/L @ 25 o C. Steps . . . Major species??? Bi 2 S 3 , H 2 O RICE is what? . . .
  • 73. Calculating Ksp pp K sp of Bi 2 S 3 , with solubility of 1.0 x 10 -15 mol/L @ 25 o C. Solubility means the amount that dissolves . R Bi 2 S 3 2 Bi 3+ + 3 S 2- I 1 .0 x 10 -15 M 0 0 C - 1.0 x 10 -15 + 2 (1.0 x 10 -15 ) 3 (1.0 x 10 -15 ) E 0 2 .0 x 10 -15 3 .0 x 10 -15 K sp = [Bi 3+ ] 2 [S 2- ] 3 = ( 2 .0 x 10 -15 ) 2 ( 3 .0 x 10 -15 ) 3 K sp = 1.1 x 10 -73 (very small number)
  • 74. Calculating Solubility from K sp pp Calculate the solubility in g /L of Cu(IO 3 ) 2 with a K sp of 1.4 x 10 -7 M . Steps . . . Major species . . . Cu(IO 3 ) 2 & H 2 O. Now do RICE R Cu(IO 3 ) 2 Cu 2+ + 2 IO 3 1- I x 0 0 C -x +x + 2 x E 0 x 2x
  • 75. Calculating Solubility from K sp pp Calculate the solubility in g /L of Cu(IO 3 ) 2 (M = 413g/mol) with a K sp of 1.4 x 10 -7 M . R Cu(IO 3 ) 2 Cu 2+ + 2 IO 3 1- I x 0 0 C -x +x + 2 x E 0 x 2x K sp = 1.4 x 10 -7 = [x][2x] 2 = 4x 3 = ??? x = 3.3 x 10 -3 mol /L = 1.4 g /L
  • 76. Relative solubilities pp K sp alone will only allow us to compare the solubility of different solids that fall apart into the same number of ions. E.g., if asked to compare the AgI , CuI and CaSO 4 solubilities , compare K sp values ( 1.5 x 10 -16 , 5.0 x 10 -12 , 6.1 x 10 -5 ) Biggest K sp is most soluble ( CaSO 4 ). We can do this ONLY because they fall apart into the same number of ions.
  • 77. Relative solubilities pp If they fall apart into different number of pieces you have to do the math. E.g ., CuS, Ag 2 S & Bi 2 S 3 See Table 15.5 p. 722. Salt K sp Solubility @ 25 o C CuS 8.5 x 10 -45 9.2 x 10 -23 mol/L Ag 2 S 1.6 x 10 -49 3.4 x 10 -17 mol/L Bi 2 S 3 1.1 x 10 - 75 1.0 x 10 - 15 mol/L Even though Bi 2 S 3 has the least K sp , it has greatest solubility! Watch for trick AP questions on this.
  • 78. Common Ion Effect pp If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve. This is Le Chatelier’s principle. This is why you must consider initial concentrations (major species)! Calculate the solubility of solid Ca F 2 (K sp = 4.0 x 10 -11 ) in a 0.025 M Na F solution. Steps. . .
  • 79. Common Ion Effect pp Ca F 2(s) solubility in 0.025 M Na F solution. (K sp = 4.0 x 10 -11 ) Major species??? CaF 2 Na 1+ F 1- H 2 O. Write RICE (+ simplifying assumptions if K sp is <<) . . . R Ca F 2(s) Ca 2+ + 2 F 1- I x 0 0.025 C -x +x 2 x E 0 x .025 + 2 x
  • 80. Common Ion Effect pp Ca F 2(s) solubility in 0.025 M Na F solution. (K sp = 4.0 x 10 -11 ) R Ca F 2(s) Ca 2+ + 2 F 1- I x 0 0.025 C -x +x 2 x E 0 x .025 + 2 x 4.0 x 10 -11 = (x)(.025 + 2x) 2 a cubic!!! But, K sp = <<, so .025 + 2x ≈ 0.025 4.0 x 10 -11 = (x)(.025) 2 x = 6.4 x 10 -8 M
  • 81. pH and solubility OH - can be a common ion. E.g., Mg(OH) 2 Mg 2+ + 2OH 1- Adding OH - shifts left , so more soluble in acid. This is how antacids work. Don’t want Mg(OH) 2 too soluble or it tears up the stomach. So, it only gets more soluble when there is acid, which it then neutralizes (shifting right to do so) and the rest of it stays insoluble until needed or excreted.
  • 82. pH and Solubility pp Other anions : If they come from a weak acid ( i.e., are effective bases) their salt is more soluble in acidic solution than in water (shifting =m). CaC 2 O 4 Ca +2 + C 2 O 4 -2 H + + C 2 O 4 -2 HC 2 O 4 - H + reduces C 2 O 4 -2 (acidic solution) so salt dissolution shifts right .
  • 83. 15.7 Precipitation & Qualitative Analysis pp Ion Product, Q =[M + ] o a [Nm - ] o b If Q > K sp a precipitate forms. If Q < K sp No precipitate. If Q = K sp at equilibrium. A solution of 750.0 mL of 4.00 x 10 -3 M Ce(NO 3 ) 3 is added to 300.0 mL of 2.00 x 10 -2 M KIO 3 . Will Ce(IO 3 ) 3 (Ksp= 1.9 x 10 -10 M ) precipitate and if so, what is the concentration of the ions? Steps.
  • 84. Precipitation & Qualitative Analysis pp 750.0 mL of 4.00 x 10 -3 M Ce(NO 3 ) 3 added to 300.0 mL of 2.00 x 10 -2 M KIO 3 . Ksp= 1.9 x 10 -10 M . Will Ce(IO 3 ) 3 ppt; if so, what is [ ] of the ions? Calculate [ions] in the mixed solution before any reaction. Then calculate Q & compare to K sp . Steps follow . . .
  • 85. Precipitation & Qualitative Analysis pp 750.0 mL of 4.00 x 10 -3 M Ce(NO 3 ) 3 + 300.0 mL of 2.00 x 10 -2 M KIO 3 . Ksp= 1.9 x 10 -10 M . ( M = mol/L) [Ce 3+ ] o = (750 mL)(4 x 10 -3 mmol/ml ) (750 mL + 300 mL) [Ce 3+ ] o = 2.86 x 10 -3 M [IO 3 1- ] o = 5.71 x 10 -3 M ( same method)
  • 86. Precipitation & Qualitative Analysis pp Will Ce(IO 3 ) 3 ppt? K sp = 1.9 x 10 -10 Ce(IO 3 ) 3 Ce 3+ + 3IO 3 1- [Ce 3+ ] o = 2.86 x 10 - 3 M [IO 3 1- ] o = 5.71 x 10 -3 M (same method) Q = (2.86 x 10 -3 )(5.71 x 10 -3 ) 3 Why isn’t [IO 3 1- ] multiplied by 3 also? We already have an actual [ ] for it. Q = 5.32 x 10 -10 > K sp = 1.9 x 10 -10 Ppt forms! What are the =m ion [ ]s?
  • 87. Precipitation & Qualitative Analysis pp What are =m ion [ ]s? [Ce 3+ ] o = 2.86 x 10 - 3 M; [IO 3 1- ] o = 5.71 x 10 -3 M Do stoich , then backtrack to =m R Ce 3+ + 3IO 3 1- Ce(IO 3 ) 3 I (2.86 x 10 -3 M )(1050 mL) (5.71 x 10 -3 M )(1050 mL) = 3.00 mmol 6.00 mmol F 1.00 mmol 0 mmol (remember mole ratios & limiting reactant) . . .
  • 88. Precipitation & Qualitative Analysis pp What are =m ion [ ]s cont. R Ce 3+ + 3IO 3 1- Ce(IO 3 ) 3 F 1 mmol 0 mmol (2 mmol) Now do =m (RICE) - “Backtrack” the reaction R Ce(IO 3 ) 3 Ce 3+ + 3IO 3 1- I x 1 mmol/1050 mL 0 C -x +x + 3x E (1 mmol + x) /(1050 mL 0 ≈ 9.52 x 10 -4 3x We know that K sp = 1.9 x 10 -10 Solve for 3x = [IO 3 1- ] = (next slide)
  • 89. Precipitation & Qualitative Analysis pp What are =m ion [ ]s cont. R Ce(IO 3 ) 3 Ce 3+ + 3 IO 3 1- I x 1 mmol/1050 mL 0 C -x +x + 3x E (1 mmol/(1050 mL) + x 9.52x10 -4 + x 3 x K sp = 1.9x10 -10 solve for x, then get 3x 1.9 x 10 -10 = (9.52 x 10 -4 )(3x) 3 = (9.52 x 10 -4 )( 27 x 3 ) x = 1.95 x 10 -3 3x = [ IO 3 1- ] = 5.84 x 10 -3 [Ce 3+ ] = 2.90 x 10 -3 (= 9.52 x 10 -4 + 1.95 x 10 -3 )
  • 90. Selective Precipitation pp A solution has 1.0 x 10 -4 M Cu 1 + & 2.0 x 10 -3 M Pb 2+ If I 1- is added which ppts 1st . . . PbI 2 (K sp 1.4 x 10 -8 ) or CuI (K sp = 5.3 x 10 -12 ) ? How much I 1- is needed to ppt each salt? Steps. Need Q < K sp for no ppt. So, when Q = K sp that is the maximum amount that can solubilize. We know [Cu 1+ ] & [Pb 2+ ] & the above K sp for both insoluble compounds, so can calculate [I 1- ] that would put [Cu 1+ ] & [Pb 2+ ] into a ppt. 1.4 x 10 -8 = [Pb 2+ ][I 1- ] 2 = (2.0 x 10 -3 )[I 1- ] 2 [I 1- ] = 2.6 x 10 -3 M with Pb 2+ 5.3 x 10 -12 = [Cu 1+ ][I 1- ] = (1.0 x 10 -4 )[I 1- ] [I 1- ] = 5.3 x 10 -8 M with Cu 1+ So, CuI will precipitate out first.
  • 91. Selective Precipitations Used to separate mixtures of metal ions in solutions. Add anions that will only precipitate certain metals at a time. Often use H 2 S because in acidic solution the less soluble sulfides like Hg +2 , Cd +2 , Bi +3 , Cu +2 , Sn +4 will precipitate first, and in basic solutions . . .
  • 92. Selective Precipitation In Basic (adding OH - ) solution, S -2 (from H 2 S) will increase so the more soluble sulfides will precipitate. (E.g., Co +2 , Zn +2 , Mn +2 , Ni +2 , Fe +2 , Cr(OH) 3 , Al(OH) 3) I.e., can regulate by changing the pH
  • 93. Z5e p. 771 Fig 15.11 Precipitation by H 2 S
  • 94. Selective precipitation Follow the steps first with insoluble chlorides (Ag, Pb, Ba) Then sulfides in acid . Then sulfides in base . Then insoluble carbonate (Ca, Ba, Mg) Alkali metals and NH 4 + remain in solution (solubility song). See, flow chart figures 15.12 p. 730 as I show you how to read them.
  • 95. Z5e Fig 15.12 p771 Selective Precipitation of Common Ions
  • 96. Z5e p. 779 Fig. 15.13 Separation of Gr 1 Ions
  • 97. 15.8 Complex Ion Equilibria pp A positively charged ion surrounded by ligands . Ligands: Lewis bases using their lone pair to stabilize the charged metal ions. Common : NH 3 , H 2 O, Cl - ,CN - , SCN - , OH 1- Memorize these! Need for net reactions. Coordination number is the number of attached ligands ( usually double the ion charge ). Cu(NH 3 ) 4 +2 has a coordination # of 4
  • 98. The addition of each ligand has its own equilibrium Usually the ligand is in large excess compared to the metal ion. And the individual K’s will be large, so we can treat them as if they go to completion . The complex ion will be the biggest ion in solution.
  • 99. Metal Ion Surrounded by Water Molecules
  • 100. Complex Ion Problem pp Calc. [ Ag 1+ ], [ Ag(S 2 O 3 ) 1- ], & [ Ag(S 2 O 3 ) 2 3- ] in mixture of 150.0 mL of 1.00 x 10 -3 M AgNO 3 & 200.0 mL of 5.00 M Na 2 S 2 O 3 Ag 1+ + S 2 O 3 2- Ag(S 2 O 3 ) 1- K 1 = 7.4 x 10 8 Ag(S 2 O 3 ) 1- + S 2 O 3 2- Ag(S 2 O 3 ) 2 3 - K 2 = 3.9 x 10 4 [ ] before rxn : [Ag 1+ ] o = (150)(10 -3 )/(150+200) = 4.29x10 -4 Ditto for [S 2 O 3 2- ] o . . . = . . . 2.86 M Since [S 2 O 3 2- ] o >> [Ag 1+ ] o and K 1 & K 2 are large , assume that both formation reactions go to completion. See “RIF” next slide. Ag 1+ is limiting & << compared to S 2 O 3 2-
  • 101. Complex Ion Problem cont. pp R Ag 1+ + 2S 2 O 3 2- Ag(S 2 O 3 ) 2 3- I 4.29 x 10 -4 M 2.86 M 0 F almost 0 2.86 - 2(4.29 x 10 -4 ) 4.29 x 10 -4 ≈ 2.86 Since there is some Ag 1+ at =m & there is Ag(S 2 O 3 ) 1- in solution we calculate their [ ]s using K 1 & K 2 K 2 = 3.9x10 4 = [Ag(S 2 O 3 ) 2 3- ] [Ag(S 2 O 3 )1- ] = 3.8 x 10 -9 M [Ag(S 2 O 3 ) 1- ][2S 2 O 3 2- ] K 1 = 7.4x10 8 = [Ag(S 2 O 3 ) 1- ] [Ag 1+ ] = 1.8 x 10 -18 M [Ag 1+ ][S 2 O 3 2- ] [Ag(S 2 O 3 ) 2 3- ] >> [Ag(S 2 O 3 ) 1- ] >> [Ag 1+ ]
  • 102. Strategies to dissolve a water-insoluble solid pp If the anion is a good base then increase solubility by adding acid E.g., Mg(OH) 2 Mg 2+ + 2OH - If the anion is not a good base, “ complex it ” by dissolving in a solution that has a ligand that forms complex ions with its cation . AgCl ( s ) + 2NH 3(aq) Ag(NH 3 ) 2 1+ ( aq ) + Cl 1- (ac) Aqua Regia (HCl + HNO 3 ) “royal water” Dissolves lots of stuff, including many sulfides.