Chapter 16 Aqueous Ionic Equilibria. Common Ion Effect ● Water dissolves many substances and often...
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Chapter 16 Aqueous Ionic Equilibri a
Chapter 16 Aqueous Ionic Equilibria. Common Ion Effect ● Water dissolves many substances and often many of these interact with each other. ● A weak acid,
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?