Buffer Capacity
amount of H+ or OH- the solution can
Ch. 15: Applications absorb without a significant change in
pH
of Aqueous Equilibria determined by the size of [HA] and [A-]
if the amounts are equal (ratio of one)
15.3 Buffer Capacity and they have large concentrations, it
will have the largest buffer capacity
the pKa of weak acid used should be
very close to the pH desired for the
buffer
Buffer Capacity Example
A buffered solution contains 0.25 M
NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl.
Calculate the pH of this solution.
NH3 + H2O ⇄ NH4+ + OH-
because the [HA] is so low in solution B, the I 0.25 0.40 0
buffer capacity is much lower C -x +x +x
this causes the change in pH after adding E 0.25-
0.25-x 0.40+x x
acid to be much larger
Example
Example
(0.40 + x)( x) ( x)(0.40)
1.8 ×10 −5 = ≈ NH4+ ⇄ NH3 + H+
(0.25 − x) (0.25)
x = 1.1×10−5 = [OH −] → pOH = 4.95 → pH = 9.05
I 0.360 M 0.290 0
C -x +x +x
What would the pH be after 0.020 mol of
OH- is added to 500.0 mL of the solution? E 0.360-
0.360-x 0.290+x x
NH4+ + OH- ⇄ NH3 + H2O 1×10 −14 (0.290 + x)( x) ( x)(0.290)
I 0.200mol 0.020 0.125 = 5.6 ×10 −10 = ≈
1.8 ×10 −5 (0.360 − x) (0.360)
C -0.020 -0.020 +0.020
x = 6.9 × 10 −10 = [ H + ] → pH = 9.16
E 0.180 0 0.145
1
Titration
used to find the concentration of a
Ch. 15: Applications solution using a solution of known
concentration
of Aqueous Equilibria analyte
solution with unknown concentration
15.4 Titrations and pH curves titrant
solution with known concentration
equivalence point is usually marked
by the end point of an indicator
Strong Acid and Strong Base
pH curve
50.0 mL of 0.200 HNO3 with 0.100 M NaOH
also called a titration curve A. no NaOH-
NaOH- 50.0 mL HNO3
plotting of pH of the analyte as a pH is only from [HNO 3]=[H+]=0.200
pH=0.699
function of the amount of titrant
added B. 10.0 mL NaOH-
NaOH- 50.0 mL HNO3
H+ + OH- H2O
millimole (mmol)
mmol)
I (50.0*0.200)=10.0 mmol (10.0*0.100)=1.00 mmol
1/1000th of a mole
C -1.00 -1.00
mmol/mL = molarity
E 9.0 mmol 0
can be more convenient than mol
[H+]=9.0mmol/(50+10)mL = 0.15 M
pH=0.82
Strong Acid and Strong Base Strong Acid and Strong Base
C. 20.0 mL NaOH and 50.0 mL HNO3 E. 100.0 mL NaOH and 50.0 mL HNO3
H+ + OH- H2O H+ + OH- H2O
I (50.0*0.200)=10.0 mmol (20.0*0.100)=2.00 mmol I (50.0*0.200)=10.0 mmol (100.0*0.100)=10.0 mmol
C -2.00 -2.00 C -10.0 -10.0
E 8.0 mmol 0 E 0 mmol Equivalence point! 0
[H+]=0.80mmol/70mL = 0.11 M pH = 0.94 [H+]=0 from HNO 3 pH = 7.00
D. 50.0 mL NaOH and 50.0 mL HNO3 F. 150.0 mL NaOH and 50.0 mL HNO3
H+ + OH- H2O H+ + OH- H2O
I (50.0*0.200)=10.0 mmol (50.0*0.100)=5.00 mmol I (50.0*0.200)=10.0 mmol (150.0*0.100)=15.0 mmol
C -5.00 -5.00 C -10.0 -10.0
E 5.0 mmol 0 E 0 mmol 5.0 mmol
[H+]=5.0mmol/100mL = 0.050 pH=1.30 [OH-]=5.0mmol/200mL = 0.025 pH=12.40
2
Strong Acid and Strong Base
Weak Acid and Strong Base
very gradual changes in pH until close to the
equivalence pt.- pH at eq. pt. = 7 50 mL of 0.10M HC2H3O2 (Ka= 1.8x10-5)
with 0.1 M NaOH
no buffering area A. no NaOH-
NaOH- 50.0 mL HC2H3O2
HC2H3O2 H+ + C2 H 3 O 2 -
I 0.10 0 0
C -x +x +x
E 0.10-
0.10-x x x
x2 x2
1.8 × 10− 5 = ≈ → x = 1.3 × 10− 3
0.10 − x 0.10
[ H + ] = x → pH = 2.87
Weak Acid and Strong Base Weak Acid and Strong Base
B. 10mL NaOH and 50 mL HC2H3O2 C. 50mL NaOH and 50 mL HC2H3O2
OH- + HC2H3O2 H2O + C2H3O2 - OH- + HC2H3O2 H2O + C2H3O2 -
I (10*0.1)=1 mmol (50*0.1)=5 mmol 0 I (50*0.1)=5 mmol (50*0.1)=5 mmol 0
C -1 -1 +1 C -5 -5 +5
E 0 4 mmol 1 mmol E 0 Equivalence point! 0 5 mmol
HC2H3O2 H+ + C2 H 3 O 2 - C2H3O2 - + H2O HC2H3O2 + OH-
OH-
I 4mmol/60mL 0 1mmol/60mL I 5mmol/100mL 0 0
C -x +x +x C -x +x +x
E (4/60)-
(4/60)-x x (1/60)+x E (5/100)-
(5/100)- x x x
1.8 × 10−5 =
x( 0.017 + x) x (0.017)
→ x = 7.1 ×10−5 → pH = 4.15 1×10 −14 x2
0.067 − x
≈
0.067
= → x = [OH − ] = 5.3 × 10−6 → pH = 8.72
1.8 × 10−5 .050
Weak Acid and Strong Base Comparing Shapes
larger change at beginning
C. 75mL NaOH and 50 mL HC2H3O2
OH- + HC2H3O2 H2O + C2H3O2 - pH at eq. pt. > 7
I (75*0.1)=7.5 mmol (50*0.1)=5 mmol 0 buffering area at half-way pt.
C -5 -5 +5
E 2.5 mmol 0 5 mmol
there are two bases to consider: which will be
most important?
Can just use the OH- to determine the pH
[OH − ] = 2.5mmol / 125mL = 0.020M → pH = 12.30
3
Weak Base and Strong Acid
Weak Base and Strong Acid
larger change at beginning
maximum buffering area at half-way pt.
pH at eq. pt. < 7
