Chapter 16 Aqueous Ionic Equilibria. Common Ion Effect ● Water dissolves many substances and often...
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Transcript of Chapter 16 Aqueous Ionic Equilibria. Common Ion Effect ● Water dissolves many substances and often...
- Slide 1
- Chapter 16 Aqueous Ionic Equilibria
- Slide 2
- Common Ion Effect Water dissolves many substances and often many of these interact with each other. A weak acid, HA, dissolves in water and produces a few H + and A - ions. What happens if another source of H + or A - is present also? What does LeChatelier's Principle suggest will happen?
- Slide 3
- Common Ion Effect Addition of a strong acid to a weak acid. HCl + HC 2 H 3 O 2 Addition of a salt containing the conjugate base. HC 2 H 3 O 2 + NaC 2 H 3 O 2 Can also do the same with weak bases.
- Slide 4
- Common Ion Effect
- Slide 5
- Buffers Solutions that contain amounts of both a weak acid and a salt containing its conjugate base are said to be buffered solutions. A buffer is a solution that resists changes in pH. pH of blood buffered at 7.4 pH of seawater buffered at 8.2
- Slide 6
- Buffers
- Slide 7
- Consider the weak acid buffer of HC 2 H 3 O 2 and C 2 H 3 O 2 - (from the salt NaC 2 H 3 O 2 ). What happens when a strong base is added? What happens when a strong acid is added? Buffers have a built-in acid/base to counteract any substance.
- Slide 8
- Henderson-Hasselbach K a = [H + ] [A - ] / [HA] Rearranges to: [H + ] = K a x [HA] / [A - ] Taking the negative natural log of both sides yields:
- Slide 9
- Buffer Capacity Capacity is the amount of acid or base the buffer can neutralize. Buffers are most effective when HA:A - ratio is 1:1. The quantity of HA and/or A- cannot be exceeded = absolute capacity. When ratio becomes 1:10 or 10:1 = practical capacity.
- Slide 10
- Buffers pH of a buffer can be calculated from H-H equation. If strong acid or base is added, then a reaction needs to be written and an s.c.e. table is constructed. Do not need to re-calculate the concentrations just plug in millimole amounts.
- Slide 11
- Blood Buffers The pH of blood must be maintained at 7.40 +/- 0.05. H 2 CO 3 / HCO 3 - system pK a = 6.1 Requires [HA] = 0.0012M and [A - ] = 0.024M Acid concentration controlled by shifting CO 2 amounts
- Slide 12
- Titrations In first semester, we titrated a known base against an unknown acid using an indicator. What if we had monitored the solution with a pH meter? Graphing the results would yield a titration curve.
- Slide 13
- Titration Curves Strong acid strong base pH at equivalence is 7.00.
- Slide 14
- Titration Curves Weak acid strong base. pH at equivalence is NOT 7.00.
- Slide 15
- Titration Curve Can perform a derivative test, which shows a single peak. At halfway point, pH = pK a.
- Slide 16
- Polyprotic Acids A polyprotic acid will show multiple peaks. Measure from end of the first to second for the second proton.
- Slide 17
- Titration Curves Curve appearance depends on how weak the acid is. Weaker the acid, the lesser the inflection point and higher equivalence pH.
- Slide 18
- Titration Curves Suitable indicator depends on what pH the equivalence occurs.
- Slide 19
- Solubility Equilibria The solubility rules in Ch. 4 predict whether or not a compound is soluble. Even an insoluble compound will dissolve slightly! PbCl 2 is insoluble per the rules, but you would NOT want to drink water that comes into contact with this solid. PbCl 2(s) Pb +2 (aq) + 2 Cl - (aq) K sp = 1.7 E-5
- Slide 20
- Solubility Equilibria K sp is called the solubility product constant. Molar solubility is the number of moles of per liter of the insoluble compound that is present in solution. Molar solubility always = x. Can calculate a K sp from knowing x. Can calculate x from a K sp. MUST be able to write the reaction of the break-up correctly!
- Slide 21
- Factors in Solubility Many other factors may affect the molar solubility of a compound. Common-Ion pH Complex-Ion formation Amphoterism
- Slide 22
- Factors in Solubility The presence of a common ion will reduce the molar solubility. For the PbCl 2 equilibrium, we could add NaCl. PbCl 2(s) Pb +2 (aq) + 2 Cl - (aq) The NaCl (s) completely dissociates to Na + (aq) and Cl - (aq). What does LeChateliers Principle tell us will happen?
- Slide 23
- Factors in Solubility pH of a solution can affect many K sp reactions. Insoluble hydroxides depend entirely on the pH. Ex) Ca(OH) 2, K sp = 6.5 E-6. A saturated solution will contain ___________ M and have a pH of __________. What is the molar solubility if the pH is 10.00?
- Slide 24
- Factors in Solubility pH can also affect the solubility if the anion is the conjugate base of a weak acid. Consider: CaF 2(s) Ca +2 (aq) + 2 F - (aq) Because F- is the conjugate of HF, the following can happen: F - (aq) + H + (aq) HF (aq) Multiply this by 2 and add to the first, yields:
- Slide 25
- Factors in Solubility Formation of a complex ion occurs when several small molecules or anions bond to the metal ion 2, 4, or 6 are common. For example, two NH 3 will bond to Ag+ ion: Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) Unlike most of the equilibria in these chapters, this K is LARGE. K f = 1.7 E+7 (p. 747). Will need to solve these in two steps.
- Slide 26
- Factors in Solubility Can then combine a K sp with a K f reaction. Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) AgCl (s) Ag + (aq) + Cl - (aq) Net = The resulting K value for the net reaction = Two possibilities for the molar solubility.
- Slide 27
- Factors in Solubility Several hydroxides are amphoteric. Most notably, Al(OH) 3. Al(OH) 3(aq) + OH - (aq) Al(OH) 4 - (aq). Others are: Cr +3, Zn +2, and Sn +2.
- Slide 28
- Amphoterism
- Slide 29
- Precipitation Calculation of Q given amounts of ions, then compare to K sp. If Q > K sp, then a ppt will form. If Q < K sp, then no ppt forms. If Q = K sp, then the solution is saturated. Reminder: if adding two solutions, then you MUST account for the dilution effect!
- Slide 30
- Precipitation Ions can be selectively precipitated based on the differences in their K sp values. BaCrO 4, K sp = 1.2 E-10 SrCrO 4, K sp = 3.5 E-5 As K 2 CrO 4 is added drop wise, which will ppt first? How effective is the separation?