AP Chemistry: Chapter 17 Notes
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Chapter 17: Applications of Aqueous Equilibria
Solutions of Acids or Bases Containing a Common Ion
Suppose we have a solution of HF (Ka = 7.2 x 10-4) and NaF. The salt breaks completely into ions because it is a strong electrolyte: NaF (s) ® Na+ (aq) + F- (aq)
How does the presence of F- in solution affect the dissociation of the acid HF? LeChatelier’s Principle tells us that F- ions will shift the acid equilibrium to the left, and reduce the dissociation of HF.
common ion effect: the shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction.
· this is the same effect that we saw already with polyprotic acids—the production of protons by the first dissociation greatly inhibits the successive dissociation steps!
Sample Exercise
Calculate the [H+] and % dissociation of HF in a solution with 1.0 M HF and 1.0 M NaF
Remember, Ka = = 7.2 x 10-4
HF / / H+ / F-Initial / 1.00M / 0 / 1.0
Change / -x / +x / +x
Equilibrium / 1.0-x / x / 1.0+x
Ka = = = 7.2 x 10-4 and so x = 7.2 x 10-4 = [H+] so pH = 3.14
% dissociation = x 100 = 0.072%
Last chapter, we calculated that a 1.0 M HF solution has a 2.7% dissociation. The effect from the common ion F- is huge!
Buffered Solutions
buffered solution: one that resists a change in its pH when either H+ or OH- are added. Many body systems are buffered, like our blood.
· contain a weak acid and its salt (HF and NaF) or
· contain a weak base and its salt (NH3 and NH4Cl)
Sample Exercise
Calculate the pH of a buffered solution containing 0.50M acetic acid (Ka = 1.8 x 10-5) and 0.50M sodium acetate.
HC2H3O2 / / H+ / C2H3O2-Initial / 0.50 / 0 / 0.50
Change / -x / +x / +x
Equilibrium / 0.50-x / x / 0.50+x
Ka = = = 1.8 x 10-5 and so x = 1.8 x 10-5 = [H+] so pH = 4.74
Sample Exercise
Calculate the pH change when 0.010 mol of solid NaOH is added to 1.0 L of the buffered solution in the example above. Compare that to the pH change of adding the NaOH to 1.0L water.
The species in solution before any reaction occurs are: Na+, H2O, HC2H3O2, C2H3O2-, and OH-
OH- is a very strong base and will pull protons off of the weak acetic acid, and this will go essentially to completion: OH- + HC2H3O2 ® H2O + C2H3O2-
before rcn: .01mol .50mol 0 .50mol
after rcn: 0 mol .49mol .01mol .51mol
Now we can solve the equilibrium problem:
HC2H3O2 / / H+ / C2H3O2-Initial / 0.49 / 0 / 0.51
Change / -x / +x / +x
Equilibrium / 0.49-x / x / 0.51+x
Ka = = = 1.8 x 10-5 and so x = 1.7 x 10-5 = [H+] so pH = 4.76
Change in pH = 4.76 – 4.74 = +.02
But if we add 0.01 mol NaOH to 1.0 L water, [OH- ] = .01, so [H+]= = 1 x 10-12, so pH = 12!! The change in pH = 12 – 7 = +5!!
· buffered solutions resist changes in pH much better than non-buffered solutions do!!
· for a buffer problem, always do the simple stoichiometry first, then do an equilibrium problem, as we always do
· how does buffering work? Remember, Ka = or, rearranging, [H+] = Ka
v the equilibrium concentration of H+, and therefore the pH, depends on the ratio of [HA]/[A-]. When OH- ions are added, HA is converted to A-, and the ratio [HA]/[A-] decreases. But, if the original amounts of HA and A- are larger than the OH- added, the change in the ratio and therefore the change in pH will be small!!
v the same logic applies when protons are added to a buffered solution of a weak acid and a salt of its conjugate base. Free H+ ions do not accumulate because they react with the conjugate base.
· we can use this form of the Ka equation to actually calculate the [H+]: [H+] = Ka In a buffered solution, we usually know [HA] and [A-], so this is an easy formula to solve. Take the negative log of both sides:
pH = pKa - log or, pH = pKa + log
Henderson-Hasselbalch equation: pH = pKa + log
v for a particular buffering system, all solutions that have the same ratio of [A-]/[HA] will have the same pH!
Sample Exercise
Find the pH of a solution with 0.75M lactic acid, HC3H5O3 (Ka = 1.4 x 10-4) and 0.25M sodium lactate.
[H+] = Ka = 1.4 x 10-4 = 4.2 x 10-4M so pH = 3.38
or, we can use Henderson-Hasselbalch:
pH = pKa + log= -log(1.4 x 10-4) + log= 3.38
Summary of Buffered Solutions:
o buffered solutions contain relatively large concentrations of a weak acid and a corresponding weak base. It can be HA and A- or B and BH+.
o when H+ is added to a buffer, it reacts to completion with the weak base: H+ + A- ® HA or H+ + B ® BH+
o when OH- is added to a buffer, it reacts to completion with the weak acid: OH- + HA ® A- + H2O or OH- + BH+ ® H2O + B
o the pH of the buffer is determined by the ratio of the concentrations of weak acid and weak base. As long as the ratio does not change by much, the pH does not change by much. This is true as long as concentrations of buffering materials are large compared with the amounts of H+ or OH- added.
Buffer Capacity
buffering capacity: represents the amount of protons or hydroxide ions a buffer can absorb without big changes in pH
Sample Exercise
How much does the pH change for each solution when 0.01 mol HCl (g) is added to
1.0L of each:
a) 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
b) 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
Use Henderson-Hasselbalch to find the initial pH: pH = pKa + log
In both cases, [HC2H3O2] = [C2H3O2-] so pH = pKa = 4.74
We find that for solution a), the new pH is 4.74—there is almost no change. For solution b), the new pH is 4.56—a change of 0.18 in pH.
For the best buffering, we want the ratio of [A-]/[HA] to remain constant; [A-] = [HA] is the best situation!
· since pH = pKa + log, that gives pH = pKa + log(1) = pKa. So the pKa of the weak acid to be used in the buffer should be as close as possible to the desired buffer pH!!
Sample Exercise
We need a solution buffered at pH 4.30 and have the following acids and their sodium salts. Calculate [HA]/[A-] required for each. Which system is the best?
We need [H+] to be = 10-4.30 = 5.0 x 10-5 M. Use the formula [H+] = Ka to solve for the ratio in each case, knowing Ka: ratio of [HA]/[A-]
a) chloroacetic acid (Ka = 1.35 x 10-3) 3.7 x 10-2
b) propanoic acid (Ka = 1.3 x 10-5) 3.8
c) benzoic acid (Ka = 6.4 x 10-5) 0.78 **best system**
d) hypochlorous acid (Ka = 3.5 x 10-8) 1.4 x 103
Titrations and pH Curves
pH curve or titration curve: the progress of an acid-base titration monitored by plotting pH versus the amount of titrant added
· strong acid-strong base titration: the net ionic equation is always H+ + OH- ® H2O It is easier to use millimole units (1/1000 of a mole; mmol)
molarity =
Let’s titrate 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH:
a) no NaOH added: [H+] = [HNO3] = 0.200 M; pH = .699
10.0 mL NaOH added: 1 mmol of NaOH has been added, so 1 mmol of H+ has been used up. [H+] = = 0.15 M; pH = 0.82
b) 20.0 mL NaOH added: 2 mmol of NaOH has been added, so 2 mmol of H+ has been used up. [H+] = = 0.11 M; pH = 0.942
c) 50.0 mL NaOH added: same process as above; pH = 1.301
d) 100.0 mL NaOH added: at this point, we have added 10 mmol and we originally had 10 mmol of acid. We are at the equivalence point of the titration. There is no H+ or OH- besides what comes from H2O, so the pH = 7.
e) 150.0 mL NaOH added: now there is excess NaOH—15 mmol OH- - 10 mmol H+ = 5 mmol OH- [OH-] = = 0.025 M; pOH = 1.60 so pH = 12.40
f) 200.0 mL NaOH added: same process as above; pH = 12.60
v Check text and notes for a titration curve of a strong acid by a strong base!
v Check text and notes for a titration curve of a strong base by a strong acid!
· weak acid-strong base titration: these are the same as buffer problems—first do the simple stoichiometry to determine how much acid remains, then do an acid equilibrium problem.
Let’s titrate 50.0 mL of 0.10 M acetic acid (Ka = 1.8 x 10-5) with 0.10 M NaOH:
a) no NaOH added: this is a typical equilibrium problem, where Ka = = 1.8 x 10-5 and so pH = 2.87
b) 10.0 mL NaOH added: 1 mmol NaOH reacts with 1 mmol HC2H3O2 and forms 1 mmol C2H3O2-. [HC2H3O2] = = 0.067 M [C2H3O2-] = = 0.0167 M Ka = = 1.8 x 10-5 = from this we get x = [H+] = 7.2 x 10-5 so pH = 4.14
**can do this with Henderson-Hasselbalch equation as well!! It’s quicker and easier!!*
pH = pKa + log so pH = 4.74 + log (.0167/.067) = 4.14
c) 25.0 mL NaOH added: 2.5 mmol of NaOH have been added, so 2.5 mmol of acetic acid are reacted with, and 2.5 mmol are left. Also, 2.5 mmol of acetate ion are formed. [HC2H3O2] = [C2H3O2-] = 0.033 M. Plugging into Ka, we get Ka = 1.8 x 10-5 = pH = 4.74 ··We are halfway to the equivalence point (half the acid is neutralized). At this point, we have seen that Ka = [H+] and so pH = pKa.
d) 40.0 mL NaOH added: follow the same process; pH = 5.35
e) 50.0 mL NaOH added: all of the HC2H3O2 is reacted, and we have 5.0 mmol C2H3O2-. But remember, this is the conjugate base of a weak acid, and it reacts with water: C2H3O2- + H2O HC2H3O2 + OH- Kb = = 5.6 x 10-10 [C2H3O2-] = = 0.05 M so Kb = 5.6 x 10-10 = x = [OH-] = 5.3 x 10-6 M therefore [H+] = 1.9 x 10-9 so pH = 8.72 **note that the pH at the equivalence point of a weak acid with a strong base is always > 7. This is because of the conjugate base in solution!
f) 60.0 mL NaOH added: we have excess OH-. OH- is a much stronger base than C2H3O2-, and will actually inhibit the reaction of C2H3O2- with water. So the pH is dominated by the OH- in solution. [OH-] = = 9.1 x 10-3 therefore [H+] = 1.1 x 10-12 and pH = 11.96
v See text or notes for the titration curve of a weak acid with a strong base. Note that the equivalence point is Not a vertical line. Also note the buffering effect when [HA] = [A-], at the half-equivalence point! Also, note that the pH at the equivalence point is NOT equal to 7. versus a strong acid.
Sample Exercise: Calculate Ka
A monoprotic weak acid has an unknown Ka. 2.0 mmol of the solid acid is dissolved in 100.0 mL of water and titrated with 0.0500 M NaOH. After 20.0 mL of NaOH has been added, the pH = 6.00. What is the Ka?
First, do a simple stoichiometry: 2.0 mmol HA + 1.0 mmol NaOH = 1 mmol H2O and 1 mmol HA left. Also, 1 mmol of A- is formed.
This is halfway to the equivalence point, so [H+] = Ka and pH = pKa. If the pH = 6.00 then the pKa = 6.00 and Ka = 1.0 x 10-6.
· weak base-strong acid titration: this is the same process as for a weak acid + strong base except that Kb must be considered, not Ka.
v See text for a typical titration curve
Acid-Base Indicators
There are two ways to find the equivalence point of an acid-base titration:
1. Use a pH meter to monitor pH and then plot it vs titrant added
2. Use an acid-base indicator which changes color at the end point. **The equivalence point, defined by stoichiometry, is not always the same as the end point (where the indicator changes color)!! But we can choose an indicator so that the end point and equivalence point are very close.
· Most common indicators are complex molecules that are actually acids themselves, HIn. They exist in one color in the protonated form, HIn, and another color in the base form In-.
· For example, let’s say that a weak acid indicator, HIn, has a Ka = 1.0 x 10-8 and is red in its acid form and blue in its basic form: