BOC216: 2012
Additional pH problems and solutions
1.
Acetic acid has a pKa of 4,8. How many ml of 0,1 M acetic acid and 0,1 M
sodium acetate are required to prepare 1 litre of 0,1 M buffer solution
having a pH of 5,8?
Asynsuur het ‘n pKa van 4,8. Hoeveel ml 0,1 M asynsuur en 0,1 M
natriumasetaat word benodig om 1 liter van ‘n 0,1 M bufferoplossing met ‘n
pH van 5,8 voor te berei?
Substitute the values for the pKa and the desired pH into the HendersonHasselbalch equation (Eqn 2.11) to get the ratio of conjugate base to weak acid:
pH
= pKa + log {[A-]/[HA]}
5,8 = 4,8 + log ([Acetate]/[Acetic acid])
5,8 – 4,8 = log ([Acetate]/[Acetic acid])
1,0
= log ([Acetate]/[Acetic acid])
Antilog both sides:
10
= [Acetate]/[Acetic acid]
 [Acetate]
= 10
[Acetic acid]
1
So [Acetate] = 10[Acetic acid]
For each volume of acetic acid, 10 volumes of acetate must be added (making a
total of 11 volumes of the two ionic species). Now multiply the proportion of each
component by the desired volume (1000 ml):
Acetic acid needed:
1/11 x 1000 ml = 91 ml
Acetate needed:
10/11 x 1000 ml = 909 ml
NOTE: when the ratio of {conjugate base] to [conjugate acid] is 10:1, the pH is
exactly one unit above the pKa and vice versa.
2.
What is the pH of a mixture of 5 ml of 0,1 M sodium acetate and 4 ml of 0,1
M acetic acid?
Wat is die pH van ‘n mengsel van 5 ml 0,1 M
natriumasetaat en 4 ml 0,1 M asynsuur?
We now have 5 + 4 = 9 ml of the mixture and the new (diluted) concentrations are:
[Acetate]
= 5/9 x 0,1 M = 0,0556 M
[Acetic acid] = 4/9 x 0,1 M = 0,0444 M
The new [ ] values can be substituted (with the value for pKa from Table 2.6) into
the Henderson-Hasselbalch equation or, since it is really just the ratio of [A -] to
[HA] and the original concentrations are equal, we can substitute the volumes :
pH
= pKa + log {[A-]/[HA]}
= 4,76 + log (5/4)
= 4,76 + log (1,25)
= 4,76 + (+0,097)
= 4,86
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3.
How does the pH change on adding 1 ml of 0,1 M HCl to the above
mixture?
Hoe verander die pH indien 1 ml 0,1 M HCl by bg. Mengsel gevoeg word?
Addition of HCl provides H+ which combines with acetate to give acetic acid
(equilibrium shifts to left). Therefore [Acetate] decreases, [Acetic acid]
increases and ratio and pH changes.
The new volume of the mixture is now 9 + 1 = 10 ml
Calculate new concentrations, taking into account dilution and increase or decrease
caused by addition of H+ :
[Acetate]
= (5/10 x 0,1 M) – (1/10 x 0,1 M) = 0,05 – 0,01 = 0,04 M
[Acetic acid] = (4/10 x 0,1 M) + (1/10 x 0,1 M) = 0,04 + 0,01 = 0,05 M
The new [ ] values can be substituted (with the value for pKa from Table 2.6) into
the Henderson-Hasselbalch equation:
pH
= pKa + log {[A-]/[HA]}
= 4,76 + log (0,04/0,05)
= 4,76 + log (0,8)
= 4,76 + (-0,097)
= 4,66
pH decreases from 4,86 to 4,66
4.
What is the pH of the following solutions
a)
0,35 M HCl
pH
5.
a)
= -log[H+]
= -log (0,35)
= 0,456
The weak acid HA is 2% ionised in a 0,2 M solution.
What is the Ka for this acid?
The acid is 2% ionised. From the equilibrium reaction HA  H+ + A- we can
conclude that 98% is in the form of HA and 2% is in the form of each of the
products (1 mole of HA yields 1 mole of H+ and 1 mole of A-)
First calculate concentrations of HA, H+ and A- :
[HA] = 98/100 x 0,2 M = 0,196 M
[H+] = 2/100 x 0,2 M = 0,004 M
[A-] = 2/100 x 0,2 M = 0,004 M
Using the eqn:
2
 Ka = (0,004 x 0,004) / 0,196
= 8,16 x 10
b)
-5
M
What is the pH of this solution?
First calculate pKa from the Ka calculated in part a:
pKa = -logKa
= -log(8,16 x 10-5)
= 4,088
and then use the H-H eqn:
pH
= pKa + log {[A-]/[HA]}
= 4,088 + log (0,004/0,196)
= 4,088 + log (0,02041)
= 4,088 – 1,690
= 2,398
6.
What is the pH of the following buffer mixtures?
a)
1 M acetic acid plus 0,5 M sodium acetate:
Get pKa value of 4,76 for acetic acid from the table in the textbook and use the
H-H eqn:
pH
= pKa + log {[A-]/[HA]}
= pKa + log {[acetate]/[acetic acid]}
= 4,76 + log (0,5/1)
= 4,76 + log 0,5
= 4,76 – 0,3
= 4,46
b)
0,3 M phosphoric acid H3PO4 plus 0,8 M KH2PO4 :
Phosphoric acid (H3PO4 ) is polyprotic and can lose 3 protons so we must make sure
that the correct pKa value is chosen. Since the conjugate base is KH2PO4 the
correct pKa value from Table ... is 2,14.
Use the H-H eqn:
pH
= pKa + log {[A-]/[HA]}
= pKa + log {[H2PO4-]/[H3PO4]}
= 2,14 + log (0,8/0,3)
= 2,14 + log 2,6667
= 2,14 + 0,43
= 2,57
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7.a)
Suppose you wanted to make a buffer of exactly pH 7,00 using KH2PO4
and Na2HPO4. If you had a solution of 0,1 M KH2PO4 , what concentration
of Na2HPO4 would you need?
First identify the acid and it’s conjugate base and then retrieve the correct pKa
value from the table in the text book:
KH2PO4 is the acid (has 2 x H) and Na2HPO4 is the conjugate base.
From Table ... it is clear that we must use the 2nd pKa value (6,86).
Use the H-H eqn:
pH
= pKa + log {[A-]/[HA]}
= pKa + log {[HPO42-]/[H2PO4-]}
7,00 = 6,86 + log {[HPO42-]/0,1)
7,00 – 6,86 = log {[HPO42-]/0,1)
0,14 = log {[HPO42-]/0,1)
Antilog both sides:
1,38 = [HPO42-]/0,1
1,38 x 0,1 = [HPO42-]
 [HPO42-] = 0,138 M
Must use a Na2HPO4 concentration of 0,138 M.
b)
Assume you wish to make a buffer at the same pH using the same
substances, but want the total phosphate molarity ([HPO42-] + [H2PO4-]) to
equal 0,3. What concentrations of KH2PO4 and Na2HPO4 would you use?
From part a) you were given a concentration for KH2PO4 of 0,1 M and calculated a
concentration of 0,138 M. for Na2HPO4 . Added together this results in a total
buffer concentration of 0,238 M. However, we need a buffer concentration of
0,3. In order to solve this problem we simply multiply the acid and base
concentrations by the same factor that the total buffer concentration must
increase by:
Existing buffer concentration = 0,238 M
New buffer concentration = 0,3 M
Factor = 0,3/0,238 = 1,26
 new [KH2PO4 ] = 0,1 M x 1,26 = 0,126 M
and new [Na2HPO4 ] =0,138 M x 1,26 = 0,174 M
8.
A 500 ml sample of a 0,100 M formate buffer, pH 3,75, is treated with 5 ml of
1,00 M KOH. What is the pH following this addition?
Inspection of Table .... reveals that the pKa value of the formic acid/formate pair
is 3,75, the same as the given pH. This means that the ratio of [formate] to
[formic acid] is 1 . Also, if the total [buffer] = 0,1 M and the acid and base are
present in equal amounts it follows that the individual concentrations of formate
and formic acid are 0,05 M each (0,05 + 0,05 = 0,1).
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Addition of OH- (in the form of KOH) is going to change the pH by changing the
ratio of formate to formic acid as follows:
the OH- will combine with formic acid to form water and formate
the [formate] will therefore increase and
the [formic acid] will decrease
The new concentrations will increase or decrease by the same number of moles of
OH- added to the solution and can be calculated on a mole basis as follows:
New moles formate
= (500/1000 x 0,05M) + (5/1000 x 1 M)
= 0,025 moles + 0,005 moles
= 0,03 moles
New moles formic acid
= (500/1000 x 0,05M) - (5/1000 x 1 M)
= 0,025 moles - 0,005 moles
= 0,02 moles
It is not necessary to calculate the molar concentrations as only the ratio is used
in the H-H eqn (i.e. you will get the same answer if you use the concentrations):
pH
9.
= pKa + log {[A-]/[HA]}
= 3,75 + log (0,03/0,02)
= 3,75 + log (1,5)
= 3,75 + 0,176
= 3,93
You need to make a buffer of pH 7,0 and you can choose from the weak
acids in Table ...... on p ..... Explain your choice.
The best choice would be a mixture of H2PO4- and HPO42- which has a pKa of 6,86.
This value is the closest to pH 7,0 and maximal buffering action is observed at or
near the pKa value.
10.
Describe the preparation of 2,0 litres of 0,100 M glycine buffer, pH 9,0, from
glycine (MW = 75,07 g/mol) and 1,00 M NaOH. The pKa of glycine is 9,6.
All of the glycine is derived from the solid glycine so we can calculate the amount
of glycine to use as follows:
2 l x 0,1 mol/l x 75,07 g/mol = 15,014 g
To get to pH 9,0 we must add enough OH- ions to convert some of the zwitterionic
glycine (gly±) to the conjugate base form (gly-). To get the relative concentrations
of each we use the H-H eqn:
pH
= pKa + log {[A-]/[HA]}
5
9,0 = 9,6 + log {[gly-]/[ gly±]}
9,0 – 9,6
= log {[gly-]/[ gly±]}
-0,6 = log {[gly-]/[ gly±]}
antilog both sides….
0,251
= [gly-]/[ gly±] so intuitively:. a 1:4 ratio for base:acid or 0,2M to 0,8M
Mathematically:
However, we still have two unknowns, so we must rather refer to the one in terms
of the other:
[gly-] = 0,251[gly±]
The total buffer concentration is 0,1 M so we can calculate [gly±] as follows:
[gly-] + [gly±] = 0,1
Substituting…
0,251[gly±] + [gly±] = 0,1
1,251[gly±] = 0,1
 [gly±]
= 0,1/1,251
±
 [gly ]
= 0,08 M
Substituting….
[gly-] + [gly±] = 0,1
[gly-] + 0,08 = 0,1

[gly-] = 0,1 - 0,08

[gly-] = 0,02 M
We must therefore add NaOH to a final concentration of 0,02 M. For 2 l this is
equal to 0,04 mole (from 0,02 mol/l) which can be obtained by adding:
(0,04/1,00) x 1000 = 40 ml of the 1,00 M NaOH solution
The buffer is then made up by dissolving the 15,014 g of glycine in about 1600 ml
of distilled water, adding the 40 ml of 1,00 M NaOH and then making the volume
up to 2000 ml with more distilled water.
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