Buffer and Titration Review

Chem I | Chem II AP
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1)
What is the effect on pH of adding sodium benzoate to an aqueous solution of benzoic acid?
Why does this change occur?
2)
Calculate pH:
a)
0.150 M C6H5COOH (Ka= 6.14 x 10-5)
C6H5COOH + H2O → H3O+ + C6H5COO-
0.150 M - x
≈ 0.150
  
x
 
 
x
 
Ka =
[H3O+][C6H5COO-]
[C6H5COOH]
  6.14 x 10-5 =
x2
0.150
  x = 0.00303
pH = - log (0.00303) = 2.52
b)
0.150 M C6H5COOH with 0.350 M C6H5COONa
C6H5COOH + H2O → H3O+ + C6H5COO-
0.150 M - x
≈ 0.150
  
x
 
 
0.350 M + x
≈ 0.350
Ka =
[H3O+][C6H5COO-]
[C6H5COOH]
  6.14 x 10-5 =
(0.350)(x)
0.150
  x = 2.63 x 10-5
pH = - log (2.63 x 10-5) = 4.58

note: Since Na+ is a spectator ion, it is ommited from the equation and subsequent calculations.

3)
A solution is prepared from 0.150 mole of CH3COOH, 0.010 mole of CH3COONa, and enough water to make a total volume of 1.00 liter. Calculate the [H3O+].
Ka for CH3COOH = 1.8 X 10-5
(Write your answer in scientific notation using the e symbol. ex: 1000 = 1e3)
CH3COOH + H2O → H3O+ + CH3COO-
0.150 M - x x0.010 M + x
≈ 0.150  ≈ 0.010
Ka =
[H3O+][CH3COO-]
[CH3COOH]
  1.8 x 10-5 =
(0.010)(x)
0.150
  x = [H3O+] = 2.7 x 10-4

note: Since Na+ is a spectator ion, it is ommited from the equation and subsequent calculations.

4)
Determine the pH of a solution containing 0.085 M NH3 and 0.247 M NH4Cl.
Kb for NH3 = 1.8 X 10-5
NH3 + H2O → NH4+ + OH-
0.085 M - x x0.247 M + x
≈ 0.085  ≈ 0.247
Kb =
[NH4+][OH-]
[NH3]
  1.8 x 10-5 =
(0.247)(x)
0.085
  x = 6.19 x 10-6
pOH = - log (6.19 x 10-6) = 5.21 pH = 14 - 5.21 = 8.79
note: Since Cl- is a spectator ion, it is ommited from the equation and subsequent calculations.
5)
What must be the ratio of HAc to NaAc to prepare a buffer whose pH = 4.81?
: 1
CH3COOH + H2O → H3O+ + CH3COO-
I prefer to use the Henderson-Hasselbach equation for this type of problem, but it can also be solved with the Ka equation.
pH = pKa + log (
[CH3COO-]
[CH3COOH]
)  4.81 = 4.74 + log (
[CH3COO-]
[CH3COOH]
)  
[CH3COO-]
[CH3COOH]
= 1.17
Since the question is asking for the ratio of HAc to NaAc the answer must be inverted:
[CH3COOH]
[CH3COO-]
= 0.86
If you would prefer to use the Ka equation: Ka =
[H3O+][CH3COO-]
[CH3COOH]
  1.8 x 10-5 =
(1.55 x 10-5)[CH3COO-]
[CH3COOH]
  
[CH3COOH]
[CH3COO-]
= 0.86

 

6)
How many grams of NaAc must be dissolved in a 0.200 M HAc solution to make a 400. mL buffer solution with a pH of 4.56?
g
CH3COOH + H2O → H3O+ + CH3COO-
Again, I prefer to use the Henderson-Hasselbach equation for this type of problem, but it can also be solved with the Ka equation.
pH = pKa + log (
[CH3COO-]
[CH3COOH]
)  4.56 = 4.74 + log (
[CH3COO-]
0.200 M
)  -0.18 = log (
[CH3COO-]
0.200 M
)  
[CH3COO-]
0.200 M
= 0.66   [CH3COO-] = 0.132
0.132 M =
x mol
0.400 L
  x = 0.0529 mol
0.0529 mol x
82.0 g NaAc
1 mol
= 4.3 g

 

7)
Which of the following acids (and conjugate base salts) would be the most useful in preparing a buffer solution with a pH of 4.58?
Acid A: Ka = 1.8 X 10-5
Acid B: Ka = 1.36 X 10-3
Acid C: Ka = 1.6 X 10-10

The pKa of Acid A is closest to the pH needed.

8)
A solution contains 0.300 moles of HAc and 0.200 moles of KAc in a total volume of 500. mL.
a)
Determine pH of this solution.
HC2H3O2+H2OH3O++C2H3O2-
0.300 mol / 500. mL     0.200 mol / 500. mL
0.600 M - x   x 0.400 M + x
Ka =
[H3O+][C2H3O2-]
[HC2H3O2]
  1.8 x 10-5 =
(x)(0.400 M)
(0.600 M)
  x = [H3O+] = 2.7 x 10-5
pH = 4.56
b)
Determine pH after the addition of 0.100 mole of H3O+.
The H3O+ that is added will react with the C2H3O2- to produce HC2H3O2. That is why the 0.1 mol is subtracted from the H3O+ and C2H3O2- and added to the HC2H3O2.
HC2H3O2+H2OH3O++C2H3O2-
0.3 + 0.1 = 0.4 mol   0.1 - 0.1 = 0 mol 0.2 - 0.1 = 0.1 mol
0.400 mol / 500. mL     0.100 mol / 500. mL
0.800 M - x   x 0.200 M + x
Ka =
[H3O+][C2H3O2-]
[HC2H3O2]
  1.8 x 10-5 =
(x)(0.200 M)
(0.800 M)
  x = [H3O+] = 7.2 x 10-5
pH = 4.14
c)
Determine pH after the addition of 0.100 mole of OH-.
The OH- that is added will react with the HC2H3O2 to produce C2H3O2- and water. That is why the 0.1 mol is subtracted from the HC2H3O2 and added to the C2H3O2-.
HC2H3O2+H2OH3O++C2H3O2-
0.3 - 0.1 = 0.2 mol     0.2 + 0.1 = 0.3 mol
0.200 mol / 500. mL     0.300 mol / 500. mL
0.400 M - x   x 0.600 M + x
Ka =
[H3O+][C2H3O2-]
[HC2H3O2]
  1.8 x 10-5 =
(x)(0.600 M)
(0.400 M)
  x = [H3O+] = 1.2 x 10-5
pH = 4.92
d)
How many mL of 6.00 M NaOH must be added so that the pH will be equal to the pKa?
In order for the pH to be equal to the pKa the concentrations of the acid and base have to be equal. In order for the concentrations to be equal, 0.05 mol of NaOH must be added.
HC2H3O2+H2OH3O++C2H3O2-
0.3 - 0.05 = 0.25 mol     0.2 + 0.05 = 0.25 mol
0.250 mol / 500. mL     0.250 mol / 500. mL
0.500 M - x   x 0.500 M + x
Ka =
[H3O+][C2H3O2-]
[HC2H3O2]
  1.8 x 10-5 =
(x)(0.500 M)
(0.500 M)
  x = [H3O+] = 1.8 x 10-5
pH = pKa = 4.74
To find the mL of NaOH: 6 M =
0.05 mol
x L
  x = 0.00833 L = 8.33 mL
9)
17.8 mL of a 0.344 M H2SO4 solution is required to completely neutralize 20.0 mL of a KOH solution. Calculate the concentration of the KOH solution.
M
MaVa = MbVb
(0.688 M)(17.8 mL) = (x)(20.0 mL)  x = 0.612 M
The concentration of H2SO4 is doubled because there are 2 H+ ions for every H2SO4 molecule.
10)
Calculate the pH after the following volumes of 0.300 M NaOH have been added to 40.00 mL of 0.600 M HCl. [Consider each addition independent of the others].
Before starting the titration, you should calculate the initial number of moles of HCl in your Erlenmeyer flask.
0.600 M =
x mol
0.040 L
  x = 0.0240 mol H+ initially
a)
0.00 mL
The only thing in your flask initially is 40 mL of 0.600 M HCl so you just have to take the -log of the H+ concentration.
pH = - log (0.600 ) = 0.22
b)
5.00 mL
Now that there is H+ and OH- in the flask, we have to determine which one is in excess and how much excess there is.
0.300 M =
x mol
5.00 mL
  x = 0.0015 mol OH-
0.024 mol H+ initially
0.0015 mol OH- added
0.0225 mol H+ excess
0.0225 mol excess H+
0.045 L total volume
= 0.500 M H+  pH = - log (0.500) = 0.30
c)
40.00 mL
Now that there is H+ and OH- in the flask, we have to determine which one is in excess and how much excess there is.
0.300 M =
x mol
40.00 mL
  x = 0.0012 mol OH-
0.024 mol H+ initially
0.0012 mol OH- added
0.0012 mol H+ excess
0.0012 mol excess H+
0.08 L total volume
= 0.150 M H+  pH = - log (0.150) = 0.82
d)
80.00 mL
Now that there is H+ and OH- in the flask, we have to determine which one is in excess and how much excess there is.
0.300 M =
x mol
80.00 mL
  x = 0.0024 mol OH-
0.024 mol H+ initially
0.024 mol OH- added
0 mol either in excess
There is no excess H+ or OH- which means their concentrations are equal and the pH = 7
e)
80.50 mL
Now that there is H+ and OH- in the flask, we have to determine which one is in excess and how much excess there is.
0.300 M =
x mol
80.50 mL
  x = 0.02415 mol OH-
0.024 mol H+ initially
0.02415 mol OH- added
0.00015 mol OH- excess
0.00015 mol excess OH-
0.1205 L total volume
= 0.0012 M OH-  pOH = - log (0.0012) = 2.9  pH = 14 - 2.9 = 11.1
f)
90.00 mL
Now that there is H+ and OH- in the flask, we have to determine which one is in excess and how much excess there is.
0.300 M =
x mol
90.00 mL
  x = 0.027 mol OH-
0.024 mol H+ initially
0.027 mol OH- added
0.003 mol OH- excess
0.003 mol excess OH-
0.13 L total volume
= 0.023 M OH-  pOH = - log (0.0023) = 1.64  pH = 14 - 1.64 = 12.36
11)
If 25.00 mL of Ca(OH)2 requires 18.34 mL of 0.100 M HCl to reach the equivalence point, what is the concentration of the Ca(OH)2?
M
MaVa = MbVb
(0.100 M)(18.34 mL) = (x)(25.00 mL)  x = 0.0734 M OH-
Ca(OH)2 → Ca+2 + 2 OH-
Since there are 2 OH- for every Ca(OH)2, the concentration would need to be halved.
0.0734 M OH- = 0.037 M Ca(OH)2
12)
Calculate the pH at the equivalence point for the titration of:
a)
50 mL of 0.10 M NH3 with 0.10 M HCl
At the equivalence point all the NH3 has reacted with the H+ added and the only thing that is left to react is the NH4+ with water.
Since there were initially 0.005 mol NH3, that means that 0.005 mol HCl was added and 0.005 mol NH4+ was produced.
Since the solutions are equimolar 50 mL of NH3 will need 50 mL of HCl to be neutralized, so the total volume is 100 mL of solution.
NH4++ H2O → NH3 + H3O+
0.005 mol / 0.10 L   
0.05 M - x + x+ x
Ka =
[NH3][H3O+]
[NH4+]
  5.6 x 10-10 =
x2
0.05
  x = 5.29 x 10-6  pH = 5.28
Make sure to use the Ka of NH4+, not the Kb of NH3
b)
50 mL of 0.010 M HAc with 0.10 M NaOH
At the equivalence point all the HAc has reacted with the OH- added and the only thing that is left to react is the Ac- with water.
Since there were initially 0.0005 mol HAc, that means that 0.0005 mol OH- was added and 0.0005 mol Ac- was produced.
Since the HAc is 10 times less concentrated than the OH- that means that it will require 10 times as much volume to neutralize. 50 mL of HAc will react with 5 mL of OH- for a total of 55 mL.
Ac-+ H2O → HAc + OH-
0.0005 mol / 0.055 L   
0.00909 M - x + x+ x
Kb =
[HAc][OH-]
[Ac-]
  5.6 x 10-10 =
x2
0.00909
  x = 2.26 x 10-6  pOH = 5.65  pH = 8.35
Make sure to use the Kb of Ac-, not the Ka of HAc.
13)
A solution containing 100.0 mL of 0.135 M HAc is being titrated with 0.540 M NaOH. Find pH at:
Before starting the titration, you should calculate the initial number of moles of HAc in your Erlenmeyer flask and the volume of NaOH required to neutralize it.
0.135 M =
x mol
0.100 L
  x = 0.0135 mol HAc initially
MaVa = MbVb (0.135 M)(100.0 mL) = (0.540 M)(x mL) x = 25 mL of NaOH required to neutralize
a)
the beginning
At the beginning, the only thing in the flask is the 0.135 M HAc, so the pH is based on its Ka.
HAc+ H2O → Ac- + H3O+
0.135 M - x + x+ x
Ka =
[Ac-][H3O+]
[HAc]
  1.8 x 10-5 =
x2
0.135
  x = 0.0016 M  pH = 2.81
b)
halfway to the equivalence point
Halfway to the equivalence point is the midpoint of the titration. At that point the pH is equal to the pKa.
We previously calculated that the equivalence point would require 25 mL, so the midpoint must require 12.5 mL of NaOH. The total volume will be 112.5 mL.
0.540 M =
x mol
0.0125 L
  x = 0.00675 mol OH- added
HAc+ H2O → H3O+ + Ac-
0.0135 mol - 0.0675 mol =   0.00675 mol
0.00675 mol / 0.1125 L =   0.00675 mol / 0.1125 L =
0.060 M x0.060 M
Ka =
[Ac-][H3O+]
[HAc]
  1.8 x 10-5 =
(0.060)(x)
(0.060)
  x = 1.8 x 10-5  pH = 4.74
c)
at the equivalence point
At the equivalence point, the only thing present to react in the container is the Ac- and water so the equation is changed to reflect that.
We previously calculated that the equivalence point would require 25 mL, so the total volume will be 125 mL.
0.540 M =
x mol
0.025 L
  x = 0.0135 mol OH- added
HAc+ H2O → H3O+ + Ac-
0.0135 mol - 0.0135 mol =   0.0135 mol / 0.125 L =
0 mol excess HAc  0.108 M
Ac-+ H2O →HAc + OH-
0.108 M - x xx
Kb =
[HAc][OH-]
[Ac-]
  5.6 x 10-10 =
x2
0.108
  x = 7.8 x 10-6  pOH = 5.11  pH = 8.89
d)
5.00 mL past the equivalence point
At this point, 30 mL of NaOH have been added for a total volume of 130 mL. .
There should be excess OH- in the flask which will contriube significantly towards the pH. The Ac- that is produced will produce an insignificant amount of OH-
0.540 M =
x mol
0.030 L
  x = 0.0162 mol OH- added
HAc+ H2O → H3O+ + Ac-
0.0135 mol - 0.0162 mol =   0.0135 mol
0.0027 mol excess OH-   not a sig.
0.0027 mol / 0.130 L =   source of OH-
0.0208 M OH-
pOH = - log [OH-] = - log (0.0208) = 1.68  pH = 14 - 1.68 = 12.32
14)
A solution containing 2.049 g of a weak acid required 43.88 mL of 0.1207 M NaOH to reach the equivalence point. What is the g-MW of the acid?
g/mol
0.1207 M =
x mol
.04388 L
  x = 0.0053 mol OH-
0.0053 mol OH- will require 0.0053 mol of the weak acid to neutralize.
g-MW =
mass
mol
=
2.049 g
0.0053 mol
= 386.9 g