nd
Entry Task: Jan 22 Thursday
Turn in Determine Ka Lab
Agenda
• Notes on Buffers
• In class practice
• HW: Buffer ws
pH of a salt (review)
• What is the pH of a 0.100 M Solution of a NaCN. The ka for HCN is 5.8 x
10-10.
CN-  HCN + OH-5
[1.0 x 10-14]
Kb
=
1.7
x
10
Kb=
[5.8 x 10-10]
Kb =
[OH-] [HCN]
[x2]
[CN-]
[0.100]
= 1.7 x 10-5
x2= 1.7 x 10-6 = 1.3 x 10-3 [OH-]
pOH = 2.88 or pH = 11.12
Buffers
• Resist the change in pH
• Contains acid/conjugate base
Or
• Contains base/conjugate acid
MAYHAN
Buffers
If you add OH-, the HF (acid) will neutralize the OH(base) , thus the HF level will decrease.
MAYHAN
Buffers
If an acid (H+) is added, the F− neutralize it to
form HF and water.
MAYHAN
Calculating the pH of Buffer
Common Ion way• What is the pH of a buffer that is 0.12M lactic acid- HC3H5O3 and 0.10M in sodium lactate, NaC3H5O3. The
ka: 1.4x 10-4.
HC3H5O3  H3O+ + C3H5O3-
Ka =
[H3O+] [C3H5O3-]
[HC3H5O3 ]
x= (1.4x 10-4)(0.12)
0.10
= 1.4 x 10-4
1.7 X 10-4 or pH = 3.77
[x][0.10]
[0.12]
= 1.4 x 10-4
Calculating the pH of Buffer
Alternative way- Henderson-Hasselbalch equation• What is the pH of a buffer that is 0.12M lactic acid- HC3H5O3 and 0.10M in sodium lactate, NaC3H5O3. The ka for
HCN is 1.4 x 10-4.
HC3H5O3  H3O+ + C3H5O3-
pH =
pKa + log
[base]
[acid]
pH = 1.4 x 10-4 + log
[0.10]
[0.12]
3.85 + (-0.0792)
pH = 3.77
Prepare a Buffer• How many grams of NH4Cl must be added to a 2.0 liter of 0.10 M NH3 to form a buffer whose pH is 9.00.
The kb is 1.8 x 10-5.
NH3  NH4+ + OH-
OH- = 1.0 x 10-5
pH 14-9 = pOH of 5
[x][1.0x10-5]
[0.10]
X
2.0 L
= 0.18 M
= 1.8 x 10-5
0.36 mol
(0.10)(1.8 x 10-5)
1.0 x 10-5
53.45g
1.0 mol
= x = 0.18 M
= 19.24 grams of NH4Cl
Buffer Capacity
• the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable
degree.
• It depends on the amount of acid or base from which the buffer is made.
1.0 M HC2H3O2 and 1.0 M NaC2H3O2
0.10 M HC2H3O2 and 0.1 M NaC2H3O2
Which of the solutions have a greater buffer capacity?
The larger molarity because it would resist MORE
pH Range
• For a buffer system, pick a pKa close to the desired pH.
• If you want a pH buffer system with a pH of 4, which acid/c. base par would you choose?
Phosphoric acid- H3PO4 / NaH3PO4 ka= 7.5 x 10-3
Benzoic Acid- HC7H5O6 / NaC7H5O6
Hydrofluoric Acid- HF / NaF
Ka= 6.3 x 10-5
Ka= 6.8 x 10-4
Benzoic Acid Its pKa is 4.2
Buffer Problems
Hurdles1st -Which species in problem is “acid” or “base”?
2nd Kb value for OH- or Ka for H+
3rd Setting up Ka or Kb expression correctly (1st)
4th Is there any changes in concentrations from given:
*Grams Moles  Molarity
*Mixing two different volumes (M1V1 = M2V2)
MAYHAN
1a. Calculate the pH of a buffer that is 0.100 M in NaHCO3 and 0.125M in Na2CO3. b. Calculate the pH
of a solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3.
Who is the conjugate in this reaction?
HCO3- so we use H2CO3 Ka value= 5.6x10-11
Provide the Ka expression
Ka = [H+][CO3-2]
[HCO3-]
x= (5.6 x 10-11)(0.100)
0.125
Provide the Ka express with #
Rearrange to get X by itself
5.6 x 10-11 = [x][0.125]
[0.100]
x = [H+]= 4.48 x 10-11
pH = -log(4.48 x 10-11) = 10.35
MAYHAN
1a. Calculate the pH of a buffer that is 0.100 M in NaHCO3 and 0.125M in Na2CO3. b. Calculate the pH of a
solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3.
NOTICE we have different volumes!!
120 mls
What will be the TOTAL VOLUME?
We have “diluted” our mixture so we set up a dilution problem- MAKE SURE volume units are the
same!!
(0.20M)(0.055L) =
(0.15M)(0.065L) = (x)(0.120 L)
(x) (0.120L)
9.17x10-2 M of NaHCO3 (new M)
8.125x10-2 Mof Na2CO3 (new M)
MAYHAN
1a. Calculate the pH of a buffer that is 0.100 M in NaHCO3 and 0.125M in Na2CO3. b. Calculate the pH of a
solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3.
Plug Back into Ka expression
Rearrange to get X by itself
5.6 x 10-11 = [x][8.125x10-2]
[9.17x10-2]
x= (5.6 x 10-11)(9.17x10-2)
8.125x10-2
x = [H+]= 6.32 x 10-11
pH = -log(6.32 x 10-11) = 10.20
MAYHAN
2a. Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. b. Calculate
the pH of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. Ka=
1.4 10-4
Provide the Ka express with #
1.4 x 10-4 = [x][0.11]
[0.12]
Rearrange to get X by itself
x= (1.4 x 10-4)(0.12)
0.11
x = [H+]= 1.52 x 10-4
pH = -log(4.48 x 10-11) = 3.82
MAYHAN
2a. Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. b. Calculate the pH
of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. Ka= 1.4
10-4
180mls
What will be the TOTAL VOLUME?
We have “diluted” our mixture so we set up a dilution problem- MAKE SURE volume units are the
same!!
(0.13M)(0.085L) =
(0.15M)(0.095L) = (x)(0.180 L)
(x) (0.180L)
6.14x10-2 M of lactic acid (new M)
7.92x10-2 M of sodium lactate(new M)
MAYHAN
2a. Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. b. Calculate the pH
of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. Ka= 1.4
10-4
Plug Back into Ka expression
Rearrange to get X by itself
1.4 x 10-4 = [x][7.92x10-2]
[6.14x10-2]
x= (1.4 x 10-4)(6.14x10-2)
7.92x10-2
x = [H+]= 1.09 x 10-4
pH = -log(1.09 x 10-4) = 3.96
MAYHAN