More Acid/Base Equilibria! (17.1 – 17.3)
17.1 The Common Ion Effect
- Consider the ionization of a weak acid, acetic acid: HC2H3O2(aq)  H+(aq) + C2H3O2–(aq)
- If we increase the [C2H3O2–] ions by adding NaC2H3O2, the equilibrium will shift to the left. (Le
Chatelier)
- This reduces the [H+] and raises the pH (less acidic)
- This phenomenon is called the common-ion effect.
- Common ion equilibrium problems are solved following the same pattern as other equilibrium
problems (ICE charts) EXCEPT the initial concentration of the common ion must be considered (it
is NOT zero).
Example 1: Does the pH increase, decrease, or stay the same on addition of each of the following?
(a) NaNO2 to a solution of HNO2
(b) (CH3NH3)Cl to a solution of CH3NH2
(c) sodium formate to a solution of formic acid
(d) potassium bromide to a solution of hydrobromic acid
(e) HCl to a solution of NaC2H3O2
(a) HNO2  H+ + NO2(b) CH3NH2 + H2O  CH3NH3+ + OH(c) HCHO2  H+ + CHO2(d) HBr  H+ + Br(e) C2H3O2-1 + H2O  HC2H3O2 + OH-1
Example 2: Using equilibrium constants from Appendix D, calculate the pH of the solution
containing 0.060 M KC3H5O2 and 0.085 M HC3H5O2
HC3H 5O 2 (aq)  H +1 (aq) + C3 H 5O-12 (aq)
initial
0.085
0
0.060
K a = 1.3 x 10-5
17. 2 Buffered Solutions
- A buffered solution or buffer is a solution that resists a change in pH after addition of small
amounts of strong acid or strong base.
- A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X– ) or weak base (B)
and its conjugate acid (HB+)
- Thus a buffer contains both:
An acidic species to neutralize added OH–
When a small amount of OH– is added to the buffer solution, the OH– reacts with the
acid in the buffer solution.
A basic species to neutralize added H+
When a small amount of H+ is added to the buffer solution, the H+ reacts with the base
in the buffer solution.
Composition of a Buffer - 4 ways to make a buffer solution:
1.)
2.)
3.)
4.)
Weak acid + salt of the acid
HCN and NaCN
weak acid: HCN weak base: CN-1
Weak base + salt of the base
NH3 and NH4 Cl
weak acid: NH4+1 weak base: NH3
EXCESS Weak acid + strong base
2 mol HCN + 1 mol NaOH  1 mol HCN + 1 mol NaCN + H2O
weak acid: HCN weak base: CN-1
2 mol NH4Cl + 1 mol NaOH  1 mol NH4Cl + 1 mol NH3 + NaCl
weak acid: NH4+1 weak base: NH3
EXCESS Weak base + strong acid
2 mol NH3 + 1 mol HCl  1 mol NH3 and 1 mol NH4Cl
weak acid: NH4+1 weak base: NH3
2 mol NaF + 1 mol HCl  1 mol NaF + 1 mole HF + NaCl
weak acid: HF
weak base: F-1
Example 3: Explain why a mixture of HCl and KCl does not function as a buffer, whereas a mixture
of HC2H3O2 and NaC2H3O2 does.
Buffer Capacity and pH
- Buffer capacity is the amount of acid or base that can be neutralized by the buffer before there is
a significant change in pH.
- Buffer capacity depends on the concentrations of the components of the buffer - the greater the
concentrations of the conjugate acid-base pair, the greater the buffer capacity.
- The pH of the buffer is related to Ka and to the relative concentrations of the acid and base.
Henderson-Hasselbalch equation – used for buffer solutions
(on AP equation sheet!!)
- These equations technically use the equilibrium concentrations of the acid (base) and the
conjugate base (acid).
- However, since the acid/base in the buffer is weak – the amount of the conjugate produced by
dissociation is generally small compared to the amount of the conjugate added as a salt. IF this is
true (it is for all AP buffer problems!) we do not need to do an equilibrium problem – just use the
INITIAL concentrations.
Example 4 (Example 2 again!): Using equilibrium constants from Appendix D, calculate the pH of
the solution containing 0.060 M KC3H5O2 and 0.085 M HC3H5O2
HC3H 5O 2 (aq)  H +1 (aq) + C3 H 5O-12 (aq)
K a = 1.3 x 10-5
initial
0.085
0
0.060
this is a BUFFER solution - use the Henerdson-Hasslebalch equation
use the initial concentrations !
Example 5: Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate
HCHO2 (aq)
initial
0.12
H+1 (aq) + CHO-12 (aq)
K a  1.4 x 10-4
0.11
Using the Henderson-Hasselbalch equation (and initial conc)
you would get the same answer doing a complete ICE chart
Example 6: A buffer is prepared by adding 20.0 g of acetic acid, HC2H3O2 and 20.0 g of sodium
acetate to enough water to form 2.00 L of solution.
(a) Determine the pH of the buffer
(b) Write the complete ionic equation for the reaction that occurs when a few drops of
hydrochloric acid are added to the buffer
(c) Write the complete ionic equation for the reaction that occurs when a few drops of
sodium hydroxide are added to the buffer
17.3 Acid-Base Titrations – Titration Curves
In an acid-base titration:
A solution of base (or acid) of known concentration (called standard) is added to an acid (or base).
Acid-base indicators or a pH meter are used to signal the equivalence point (when moles acid =
moles base).
The plot of pH versus volume during a titration is called a pH titration curve.
Example 7: Predict whether the equivalence point of each of the following titrations is below, above
or at pH 7:
a) NaHCO3 titrated with NaOH b) NH3 titrated with HCl
c) KOH titrated with HBr
at the equivalence point, only products are present in solution, so determine the
products of the reaction and then determine if the solution is acidic, basic or neutral
a) NaHCO3 + NaOH  Na2CO3 + H2O
b) NH3 + HCl  NH4Cl
c) KOH + HBr  KBr + H2O
Example 8: How many mL of 0.0850 M NaOH solution is required to titrate 40.0 mL of 0.0900 M
HNO3?
? mL
40.0 mL
0.0850 M
0.0900 M
NaOH + HNO3  H2O + NaNO3
1 mole
1 mole
Example 9: A 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution.
Calculate the pH of the solution after the following volumes of base solution have been added:
(a) 15.0 mL (b) 19.9 mL (c) 20.0 mL (d) 20.1 mL (e) 35.0 mL
mL
HBr
mol OH-1
(M) (V)
(.200) V
0.00300
M of
excess ion
(mol / tot vol)
0.0286 M H+1
pH
(a)
mL
mL
mol H+1
NaOH Total (M) (V)
(.200) V
20.0 15.0 35.0 0.00400
(b)
(c)
(d)
(e)
20.0
20.0
20.0
20.0
0.00398
0.00400
0.00402
0.00700
0.0005 M H+1
1 x 10-7 M H+1*
0.0005 M OH-1
0.0545 M OH-1
3.3
7.0
10.7
12.736
19.9
20.0
20.1
35.0
39.9
40.0
40.1
55.0
0.00400
0.00400
0.00400
0.00400
1.544
*When molarity is less than 10-6 we must consider the autoionization of water!
(H+ = 1.0*10-7)
Example 10: Calculate the pH at the equivalence point for titrating 0.200 M solutions of each of the
following bases with 0.200 M HBr: (a) NaOH (b) NH2OH
Ch 17 HW #1
17( 5, 6, 16, 20, 22, 26, 35, 38, 42, 44, 48)
Acid – Base Indicators
(Oops…a little more with packet #1!)
The equivalence point of an acid-base titration can be determined by measuring pH, but it can also be
determined by using an acid-base indicator which marks the end point of a titration by changing color.
Although the equivalence point (defined by the stoichiometry) is not necessarily the same as the end point
(where the indicator changes color), careful selection of the indicator can ensure that the difference between
them is negligible.
Acid-base indicators are complex molecules that are themselves, weak acids (represented by HIn).
They exhibit one color when the proton is attached and a different color when the proton is absent.
-8
Consider a hypothetical indicator, HIn, a weak acid with K =1.0x10 . It has a red color in acid and a blue
a
color in base.
+1
-1
HIn(aq)  H (aq) + In (aq)
red
blue
[H  ][In  ]
Ka =
[HIn]
Rearranging, we get
Ka
Ka
[In  ]
[In  ]
=
or
=
[H  ] [HIn]
[HIn] [H  ]
Suppose we add a few drops of indicator to an acid solution
whose pH = 1.0 ([H  ] = 1.0 x 101 )
K a 1.0 x 108
[In  ]
=
=
=
[HIn] [H  ] 1.0 x 101
10
7
=
1
10,000,000
1
[In  ]
=
10,000,000 [HIn]
This ratio shows that the predominant form of the indicator is HIn, resulting in a red solution.
As OH-1 is added (like in a titration) [H+1] decreases and the equilibrium shifts to the right, changing HIn to
In-. At some point in the titration, enough of the In- form will be present so we start to notice a color change.
It can be shown (using the Henderson-Hasselbalch equation) that for a typical acid-base indicator with
dissociation constant, K , the color transition occurs over a range of pH values given by pK ± 1.
a
a
-7
For example, bromthymol blue with K = 1.0 x 10 (pK = 7), would have a useful pH range of 7 ± 1 or
a
a
from 6 to 8.
***You want to select an indicator whose pKa value is close to the pH you want to detect (usually the
pH at the equivalence point)
Example 11: Use the following table to determine which of the following would be the best indicator to use
to indicate the equivalence point of the titrations described in Example 10.
Indicator
Methyl Yellow
Methyl Red
Bromthymol Blue
Phenolphthalein
10. a. pH at equivalence point was 7.0
10. b. pH at equivalence point was 3.52
Ka
1*10-4
1*10-5
1*10-7
1*10-9