CHEM1002
2014-N-2
November 2014
• Calculate the pH of a 0.020 M solution of lactic acid, HC3H5O3, at 25 °C. The pKa of
lactic acid is 3.86.
As lactic acid is a weak acid, [H3O+] must be calculated using a reaction
table:
HC3H5O3
H2O
H3O+
C3H5O3-
initial
0.020
large
0
0
change
-x
negligible
+x
+x
final
0.020 –x
large
x
x
The equilibrium constant Ka is given by:
𝐇𝟑 𝐎! [𝐂𝟑 𝐇𝟓 𝐎𝟑 ! ]
𝒙𝟐
Ka =
=
[𝐇𝐂𝟑 𝐇𝟓 𝐎𝟑 ]
𝟎.𝟎𝟐𝟎!𝒙
As pKa = -log10Ka, Ka = 10 3.86 and is very small, 0.010 – x ~ 0.010 and hence:
−
x2 = 0.020 × 10–3.86
or
x = 1.7 × 10-3 M = [H3O+]
Hence, the pH is given by:
pH = −log10[H3O+] = −log10(1.7 × 10-3) = 2.78
pH = 2.78
A 1.0 L solution of 0.020 M lactic acid is added to 1.0 L of 0.020 M sodium
hydroxide solution. Write the ionic equation for the reaction that occurs.
HC3H5O3(aq) + OH-(aq) à C3H5O3-(aq) + H2O(l)
Is the resulting solution acidic, basic or neutral? Give a reason for your answer.
All of the lactic acid will have reacted as the number of moles of sodium
hydroxide added equals the number of moles of lactic acid originally present.
The solution contains lactate ions, C3H5O3-(aq), which is the conjugate base. The
solution contains a weak base so is basic.
5
CHEM1002
2013-N-4
November 2013
Calculate the pH of a 0.010 M solution of aspirin at 25 °C. The pKa of aspirin
is 3.5 at this temperature.
As aspirin is a weak acid, [H3O+] must be calculated using a reaction
table:
C9H8O4
H2O
H3O+
C9H7O4-
initial
0.010
large
0
0
change
-x
negligible
+x
+x
final
0.010 –x
large
x
x
The equilibrium constant Ka is given by:
Ka =
𝐇𝟑 𝐎! [𝐂𝟗 𝐇𝟕 𝐎𝟒 ! ]
𝒙𝟐
=
[𝐂𝟗 𝐇𝟕 𝐎𝟒 ]
𝟎.𝟎𝟏𝟎!𝒙
As pKa = -log10Ka, Ka = 10 3.5 and is very small, 0.010 – x ~ 0.010 and hence:
−
x2 = 0.010 × 10–3.5
or
x = 1.8 × 10-3 M = [H3O+]
Hence, the pH is given by:
pH = −log10[H3O+] = −log10(1.8 × 10-3) = 2.8
pH = 2.8
Aspirin, C9H8O4 is not very soluble. “Soluble aspirin” can be made by reacting
aspirin with sodium hydroxide. Write the chemical equation for this reaction.
C9H8O4(s) + OH–(aq) → C9H7O4–(aq) + H2O(l)
Is a solution of “soluble aspirin” acidic or basic? Briefly explain your answer.
Basic. The C9H7O4–(aq) ion reacts with water (i.e. undergoes hydrolysis) to
generate a small amount of OH– ions. The C9H7O4–(aq) ion is a weak base, so the
following equilibrium reaction lies very much in favour of the reactants.
C9H7O4–(aq) + H2O(l)
C9H8O4(aq) + OH–(aq)
Marks
7
CHEM1002
2012-N-2
November 2012
• Solution A consists of a 0.050 M aqueous solution of benzoic acid, C6H5COOH, at
25 °C. Calculate the pH of Solution A. The pKa of benzoic acid is 4.20.
As benzoic acid is a weak acid, [H3O+] must be calculated using a reaction
table:
C6H5COOH
H+
C6H5COO-
initial
0.050
0
0
change
-x
+x
+x
final
0.050 –x
x
x
The equilibrium constant Ka is given by:
Ka =
𝐇 ! [𝐂𝟔 𝐇𝟓 𝐂𝐎𝐎! ]
𝒙𝟐
=
[𝐂𝟔 𝐇𝟓 𝐂𝐎𝐎𝐇]
𝟎.𝟎𝟓𝟎!𝒙
As pKa = -log10Ka, Ka = 10 4.20 and is very small, 0.050 – x ~ 0.050 and hence:
−
x2 = 0.050 × 10–4.2
or
x = 1.78 × 10-3 M = [H+]
Hence, the pH is given by:
pH = −log10[H+] = −log10(1.78 × 10-3) = 2.75
pH = 2.75
What are the major species present in solution A?
Ka is very small and the equilibrium lies almost completely to the left. The major
species present are water and the undissociated acid:
H2O and C6H5COOH
Solution B consists of a 0.050 M aqueous solution of ammonia, NH3, at 25 °C.
Calculate the pH of Solution B. The pKa of NH4+ is 9.24.
NH3 is a weak base so [OH-] must be calculated by considering the equilibrium:
NH3
H2O
NH4+
OH-
initial
0.050
large
0
0
change
-y
negligible
+y
+y
final
0.050 – y
large
y
y
ANSWER CONTINUES ON THE NEXT PAGE
Marks
6
CHEM1002
2012-N-2
November 2012
The equilibrium constant Kb is given by:
Kb =
𝐍𝐇𝟒 ! [𝐎𝐇 ! ]
[𝐍𝐇𝟑 ]
=
𝒚𝟐
(𝟎.𝟎𝟓𝟎!𝒚)
For an acid and its conjugate base:
pKa + pKb = 14.00
pKb = 14.00 – 9.24 = 4.76
As pKb = 4.76, Kb = 10-4.76. Kb is very small so 0.050 – y ~ 0.050 and hence:
y2 = 0.050 × 10–4.76 or y = 9.32 × 10-4 M = [OH-]
Hence, the pOH is given by:
pOH = -log10[OH-] = log10[9.32 × 10-4] = 3.03
Finally, pH + pOH = 14.00 so
pH = 14.00 – 3.03 = 10.97
pH = 10.97
What are the major species present in solution B?
Kb is very small and the equilibrium lies almost completely to the left. The major
species present are water and the unprotonated weak base:
H2O and NH3
CHEM1002
2012-N-3
November 2012
Write the equation for the reaction that occurs when benzoic acid reacts with
ammonia?
C6H5COOH(aq) + NH3(aq) → C6H5COO–(aq) + NH4+(aq)
Write the expression for the equilibrium constant for the reaction of benzoic acid with
ammonia?
𝐂𝟔 𝐇𝟓 𝐂𝐎𝐎! (𝐚𝐪) [𝐍𝐇𝟒 ! 𝐚𝐪 ]
K =
𝐂𝟔 𝐇𝟓 𝐂𝐎𝐎𝐇(𝐚𝐪) [𝐍𝐇𝟑 𝐚𝐪 ]
What is the value of the equilibrium constant for the reaction of benzoic acid with
ammonia? Hint: multiply the above expression by [H+]/[H+].
Multiplying the expression above by [H+] / [H+] gives:
K=
=
𝐂𝟔 𝐇𝟓 𝐂𝐎𝐎! (𝐚𝐪) [𝐍𝐇𝟒 ! 𝐚𝐪 ] [𝐇 ! 𝐚𝐪 ]
𝐂𝟔 𝐇𝟓 𝐂𝐎𝐎𝐇(𝐚𝐪) [𝐍𝐇𝟑 𝐚𝐪 ]
[𝐇 ! 𝐚𝐪 ] 𝐂𝟔 𝐇𝟓 𝐂𝐎𝐎! (𝐚𝐪)
𝐂𝟔 𝐇𝟓 𝐂𝐎𝐎𝐇(𝐚𝐪)
= 𝑲𝐚 ×
=
𝑲𝒃
!
[𝐇 𝐚𝐪 ][𝐎𝐇 ! 𝐚𝐪 ]
(𝟏𝟎!𝟒.𝟐𝟎 ) × 𝟏𝟎!𝟒.𝟕𝟔
(𝟏𝟎!𝟏𝟒 )
.
. [𝐇 !
𝐚𝐪 ]
[𝐍𝐇𝟒 ! 𝐚𝐪 ]
[𝐍𝐇𝟑 𝐚𝐪 ][𝐇 ! 𝐚𝐪 ]
=
𝑲𝐚 × 𝑲𝒃
𝑲𝐰
= 1.1 × 105
Answer: 1.1 × 105
What are the major species in the solution that results from adding together equal
amounts of solutions A and B?
The equilibrium strong favours products so the major species are:
C6H5CO2–(aq), NH4+(aq), H2O(l)
Marks
5
CHEM1002
2010-N-4
November 2010
• Hydrochloric acid in a healthy human stomach leads to a pH in the range 1-2. What
is the concentration of hydrochloric acid in the stomach?
As pH = -log10[H3O+(aq)] and HCl is a strong acid, these pH values correspond
to:
[H3O+(aq)] = 10-pH = 0.1 M at pH = 1
[H3O+(aq)] = 10-pH = 0.01 M at pH = 2
Answer: 0.01 – 0.1 M
The stomach also contains considerable amounts of chloride ions, the conjugate base
of hydrochloric acid, in the form of dissolved KCl and NaCl. Is this solution a
buffer? Explain your answer.
No. A buffer contains a mixture of a weak acid and its conjugate base. HCl is a
very, very strong acid and so Cl- is a very, very weak base.
If acid is added to the stomach, there is no base for it to react with and so the pH
will lower considerably.
If base is added to the stomach, it will react with the H3O+ present and the pH
will rise considerably.
This is not the action of a buffer.
Milk of magnesia is often taken to reduce the discomfort of acid stomach. A
teaspoon of milk of magnesia, containing 0.400 g of Mg(OH)2, is added to a 0.300 L
stomach solution with a pH of 1.3. What is the final pH of the solution?
The molar mass of Mg(OH)2 is (24.31 (Mg) + 2 × 16.00 (O) + 2 × 1.008 (H)) g
mol-1 = 58.326 g mol-1.
The number of moles in 0.400 g is therefore:
number of moles = mass / molar mass
= (0.400 g) / (58.326 g mol-1) = 0.00686 mol
The number of moles of OH- added to the stomach is therefore:
number of moles of OH- = 2 × 0.00686 mol = 0.0137 mol
ANSWER CONTINUES ON THE NEXT PAGE
Marks
8
CHEM1002
2010-N-4
November 2010
If the pH is originally 1.3, [H3O+(aq)] = 10-pH = 10-1.3 = 0.050 M. The number of
moles originally present in 0.300 L is therefore:
number of moles of H3O+ = concentration × volume
= (0.050 mol L-1) × (0.300 L) = 0.0150 mol
This will react with the added OH-, leaving:
number of moles of H3O+ = (0.0150 – 0.0137) mol = 0.0013 mol
Hence, the final concentration of H3O+ is
[H3O+(aq)] = number of moles / volume
= (0.0013 mol) / (0.300 L) = 0.0043 M
Finally,
pH = -log10[H3O+(aq)] = 2.4
Answer: 2.4
CHEM1002
2009-N-4
November 2009
• You have completed a number of acid/base titrations during your laboratory work.
What is the difference between the ‘end point’ and the ‘equivalence point’ in an
acid/base titration?
The endpoint is where the indicator changes colour and the reaction is
observed to be completed. The equivalence point is where equal amounts of
acid and base have been added. Ideally, the endpoint and equivalence point
should be as close to one another as possible.
How do you determine the concentration of a weak acid through titration with a
strong base? Include all necessary steps in your explanation.
Place a known volume (eg 25.00 mL) of the weak acid solution in a conical
flask and add 2 drops of a suitable indicator (such as phenolphthalein).
Titrate with a known concentration of the strong base from a burette until the
end point (the first permanent pink colour) is reached. Record the volume
used.
The equation of the reaction must be known
HnA(aq) + nOH–(aq) → An–(aq) + nH2O
This can then be used to calculate the moles of OH– used.
Hence calculate the moles of weak acid present in 25.00 mL.
Hence calculate the moles of weak acid present in 1000.00 mL (i.e. its
concentration).
How do you determine the pKa of a weak acid through titration with a strong base?
Include all necessary steps in your explanation.
The titration described above is used and the titration of the weak acid is
continued past its equivalence point with a strong base, recording the pH as you
go.
Construct a graph of mL base added vs pH. This is a titration curve.
Determine the volume of base required to reach equivalence point (where slope
of line is vertical).
Divide this value by 2 to give the half equivalence point (where [HA] = [A–]).
Read the value of the pH at the half equivalence point from the titration curve.
From Henderson-Hasselbalch equation, this is the point where pH = pKa.
Marks
6
CHEM1002
2008-N-5
November 2008
Marks
• Calculate the pH of a 0.020 M solution of Ba(OH)2.
1
Ba(OH)2 is a strong base so it will completely dissociate in solution:
Ba(OH)2(s) à Ba2+(aq) + 2OH-(aq)
As each Ba(OH)2 dissociates to make 2OH-, a 0.020 M solution has [OH-(aq)] =
0.040 M.
By definition, pOH = -log10[OH-(aq)] so pOH = -log10(0.040) = 1.40.
As pH + pOH = 14.00,
pH = 14.00 – 1.40 = 12.60
pH = 12.60
• Calculate the pH of a 0.150 M solution of HNO2. The pKa of HNO2 is 3.15.
As HNO2 is a weak acid, [H3O+] must be calculated using a reaction
table:
HNO2
H2O
H3O+
NO2-
initial
0.150
large
0
0
change
-x
negligible
+x
+x
final
0.150 –x
large
x
x
The equilibrium constant Ka is given by:
Ka =
𝐇𝟑 𝐎! [𝐍𝐎𝟐 ! ]
𝒙𝟐
=
[𝐇𝐍𝐎𝟐 ]
𝟎.𝟏𝟓𝟎!𝒙
As pKa = -log10Ka, Ka = 10 3.15 and is very small, 0.150 – x ~ 0.150 and hence:
−
x2 = 0.150 × 10–3.15 or
x = 1.03 × 10-2 M = [H3O+]
Hence, the pH is given by:
pH = −log10[H3O+] = −log10(1.03 × 10-2) = 1.987
pH = 1.987
ANSWER CONTINUES ON THE NEXT PAGE
3
CHEM1002
2008-N-5
November 2008
• Calculate the pH of a solution that is 0.080 M in acetic acid and 0.160 M in sodium
acetate. The pKa of acetic acid is 4.76.
The solution contains both a weak acid (acetic acid) and its conjugate base (the
acetate ion). This is a buffer and its pH can be calculate using the HendersonHasselbalch equation. With [acid] = 0.080 M and [base] = 0.160 M,
pH = pKa + log
[𝐛𝐚𝐬𝐞]
[𝐚𝐜𝐢𝐝]
= 4.76 + log
𝟎.𝟏𝟔𝟎
𝟎.𝟎𝟖𝟎
pH = 5.06
= 5.06
2
CHEM1002
2007-N-4
November 2007
• Solution A consists of a 0.50 M aqueous solution of HF at 25 °C. Calculate the pH of
Solution A. The pKa of HF is 3.17.
As HF is a weak acid, [H3O+] must be calculated using a reaction
table:
HF
H2 O
H3 O+
HF
initial
0.50
large
0
0
change
-x
negligible
+x
+x
final
0.50 –x
large
x
x
The equilibrium constant Ka is given by:
Ka =
[H 3 O + ][F − ]
x2
=
[HF]
0.50 − x
As pKa = -log10Ka, Ka = 10−3.17 and is very small, 0.50 – x ~ 0.50 and hence:
x2 = 0.50 × 10–3.17
or
x = 1.84 × 10-2 M = [H3O+]
Hence, the pH is given by:
pH = −log10[H3O+] = −log10[(1.84 × 10-2)] = 1.74
pH = 1.74
ANSWER CONTINUES ON THE NEXT PAGE
Marks
8
CHEM1002
2007-N-4
November 2007
At 25 °C, 1.00 L of Solution B consists of 12.97 g of lithium fluoride, LiF, dissolved
in water. Calculate the pH of Solution B.
The molar mass of LiF is 6.941 (Li) + 19.00 (F) = 25.941. Hence, the number of
moles in 12.97 g is:
number of moles =
mass
12.97
=
= 0.5000 mol
molar mass 25.941
As this is dissolved in 1.00 L, [F-] = 0.500 M.
F- is a weak base so [H3O+] must be calculated via the calculation of [OH-]
from the reaction table:
F-
H2 O
HF
OH–
initial
0.500
large
0
0
change
-x
negligible
+x
+x
final
0.500 - x
large
x
x
The equilibrium constant Kb is given by:
Kb =
[HF][OH − ]
x2
=
0.500 − x
[F − ]
As pKa + pKb = 14.00, pKb = 14.00 – 3.17 = 10..83. As pKb = -log10Kb,
Kb = 10-10.83 and is very small. Hence, 0.500 – x ~ 0.500 and hence:
x2 = 0.500 × 10–10.83 or
x = 2.72 × 10-6 M = [OH-]
Hence, the pOH is given by:
pOH = −log10[OH-] = −log10[(2.72 × 10-6)] = 5.57
Finally, as pH + pOH = 14.00, pH = 14.00 – 5.57 = 8.43
pH = 8.43
ANSWER CONTINUES ON THE NEXT PAGE
CHEM1002
2007-N-4
November 2007
Solution B (1.00 L) is poured into Solution A (1.00 L) and allowed to equilibrate at
25 °C. Calculate the pH of the final solution.
The solution consists of a weak acid (HF) and its conjugate base (F-). The
Henderson -Hasselbalch equation can be used:
 [base] 
pH = pKa + log10 

 [acid] 
[base] = [F-] = 0.500 M and [acid] = [HF] = 0.50 M. Hence,
 [base] 
 0.500 
pH = pKa + log10 
 = 3.17 + log10 
 = 3.17
 0.50 
 [acid] 
pH = 3.17
If you wanted to adjust the pH of the mixture of Solution A and
Solution B to be exactly equal to 4.00, which component in the
mixture would you need to increase in concentration?
pH needs to
increase:
increase [base]
(i.e. [LiF])
CHEM1002
2006-N-4
November 2006
Marks
• What is the pH of a 0.020 M solution of HF? The pKa of HF is 3.17.
2
As HF is a weak acid, [H3O+] must be calculated using a reaction table:
HF
H2 O
H3 O+
F–
initial
0.020
large
0
0
change
-x
negligible
+x
+x
final
0.020 – x
large
x
x
The equilibrium constant Ka is given by:
Ka =
[H 3 O + ][F − ]
x2
=
[HF]
0.020 − x
As pKa = -log10(Ka), Ka = 10−3.17 and is very small. Hence, 0.020 – x ~ 0.020 and
hence:
x2 = 0.020 × 10–3.17 or
x = 3.68 × 10-3 M = [H3O+]
Hence, the pH is given by:
pH = −log10[H3O+] = −log10[(3.68 × 10-3)] = 2.43
pH = 2.43
• What is the pH of a solution that is 0.075 M in acetic acid and 0.150 M in sodium
acetate? The pKa of CH3COOH is 4.76.
This solution containing a weak acid (CH3COOH) and its conjugate base
(CH3COO-) is a buffer and the Henderson-Hasselbalch equation can be used to
calculate the pH:
 [CH 3 COO − ] 
 [base] 
pH = pKa + log10 
=
pK
(CH
COOH)
+
log
a
3


10 
 [acid] 
 [CH 3 COOH] 
 0.150 
pH = 4.76 + log10 
 = 5.06
 0.075 
pH = 5.06
THE ANSWER CONTINUES ON THE NEXT PAGE
2
CHEM1002
2006-N-4
November 2006
• What is the pH of a 0.010 M solution of Ba(OH)2?
Ba(OH)2 is a strong base and dissociates completely according to the equation:
Ba(OH)2(aq) Ba2+(aq) + 2OH-(aq)
Hence a 0.010 M solution has [OH-(aq)] = 2 × 0.010 = 0.020 M.
As pOH = -log10([OH-(aq)], pOH = -log10(0.020) = 1.70.
In aqueous solution, pH + pOH = 14.00 so pH = (14.00 – 1.70) = 12.30
pH = 12.30
2
CHEM1002
2005-N-5
November 2005
Marks
 What is the pH of a 0.010 M solution of Ca(OH)2?
2
Ca(OH)2 is a strong base and dissociates completely according to the equation:
Ca(OH)2(aq)  Ca2+(aq) + 2OH-(aq)
A 0.010 M solution therefore has [OH-(aq)] = 0.020 M and
pOH = -log10([OH-(aq)]) = -log10(0.020) = 1.70
As pH + pOH = 14.00, pH = (14.00 – 1.70) = 12.30
pH = 12.30
 What is the pH of a 0.010 M solution of HNO2? The pKa of HNO2 is 3.15.
Nitrous acid is a weak acid so [H3O+] must again be calculated:
HNO2(aq) H2O(l)
H3O+(aq)
NO2–(aq)
initial
0.010
large
0
0
change
-x
negligible
+x
+x
final
0.010 – x
large
x
x
The equilibrium constant Ka is given by:
Ka 
[H 3 O ][NO2  ]
x2

[HNO2 ]
0.010  x
As Ka = 103.15 is very small, 0.010 – x ~ 0.010 and hence:
x2 = 0.010  10–3.15
or
x = 2.66 × 10-3 M = [H3O+(aq)]
Hence, the pH is given by:
pH = log10[H3O+(aq)] = log10[(2.66 × 10-3)] = 2.58
pH = 2.58
ANSWER CONTINUES ON THE NEXT PAGE
2
CHEM1002
2005-N-5
November 2005
 What is the pH of a solution that is 0.020 M in CH3COOH and 0.010 M in CH3CO2–?
The Ka of CH3COOH is 1.8  10–5 M.
The solution contains a mixture of a weak acid (CH3COOH) and its conjugate
base (CH3COO-) so acts as a buffer and the Henderson-Hasselbalch equation
can be used:
 [base] 

pH = pKa + log 10 
 [acid] 
with [base] = [CH3COO-(aq)] and [acid] = [CH3COOH(aq)].
As Ka = 1.8 × 10-5, pKa = -log10Ka = -log10(1.8 × 10-5) = 4.74.
Hence:
 [0.010] 
pH = 4.74 + log10 
 = 4.44
 [0.020] 
pH = 4.44
2
CHEM1002
2004-N-4
November 2004
• Solution A consists of a 0.15 M aqueous solution of nitrous acid (HNO2) at 25 °C.
Calculate the pH of Solution A. The pKa of HNO2 is 3.15.
Nitrous acid is a weak acid so [H3O+] must be calculated:
HNO2
H2O
H3O+
NO2–
initial
0.15
large
0
0
change
-x
negligible
+x
+x
final
0.15 – x
large
x
x
The equilibrium constant Ka is given by:
Ka =
[H 3 O + ][NO 2 − ]
x2
=
[HNO 2 ]
0.15 − x
As Ka = 10 3.15 is very small, 0.15 – x ~ 0.15 and hence:
−
x2 = 0.15 × 10–3.15
or
x = 1.03 × 10-2 M = [H3O+]
Hence, the pH is given by:
pH = −log10[H3O+(aq)] = −log10[(1.03 × 10-2)] = 1.99
pH = 1.99
ANSWER CONTINUES ON THE NEXT PAGE
Marks
8
CHEM1002
2004-N-4
November 2004
At 25 °C, 1.00 L of Solution B consists of 13.8 g of sodium nitrite (NaNO2) dissolved
in water. Calculate the pH of Solution B.
The formula mass of NaNO2 is (22.99 (Na) + 14.01 (N) + 2 × 16.00 (O)) = 69.
Therefore, 13.8 g corresponds to:
number of moles of NaNO2 =
mass
13.8
=
= 0.200mol
formula mass 69.0
As this is dissolved in 1.00 L, the concentration is 0.200 M.
NO2– is a weak base so [OH–(aq)] must be calculated from the equilibrium:
initial
change
final
NO2–
H2O
OH–
HNO2
0.200
-y
0.200 – y
large
negligible
large
0
+y
y
0
+y
y
The equilibrium constant Kb is given by:
[OH − ][HNO 2 ]
y2
Ka =
=
0.2 − y
[NO 2 − ]
For an acid and its conjugate base, pKa + pKb = 14.00 so:
pKb = 14.00 – 3.15 = 10.85
As pKb = 10.85, Kb = 10 10.85. Kb is very small so 0.200 – y ~ 0.200 and hence:
−
y2 = 0.200 × 10–10.85 or
y = 1.68 × 10-6 M = [OH (aq)]
−
Hence, the pOH is given by pOH = −log10[OH (aq)] = −log10[1.68 × 10-6] = 5.77
−
Finally, pH + pOH = 14 so pH = (14.00 – 5.77) = 8.23
pH = 8.23
ANSWER CONTINUES ON THE NEXT PAGE
CHEM1002
2004-N-4
November 2004
Solution B (1.00 L) is poured into Solution A (1.00 L) and allowed to equilibrate at
25 °C. Calculate the pH of the final solution.
Solution A, HNO2(aq), is 0.15 M. When 1.00 L of A is added to 1.00 L of B, a
2.00 L solution is formed. This dilution halves the concentration to 0.075 M.
Similarly, solution B, KNO2(aq), which is initially 0.200 M is diluted to 0.100 M
by the addition of solution A.
The combined solution contains a mixture of a weak acid (HNO2) and its
conjugate base (NO2-) so acts as a buffer and the Henderson-Hasselbalch
equation can be used:
⎛ [base] ⎞
pH = pKa + log 10 ⎜⎜
⎟⎟ with [base] = [NO2-(aq)] and [acid] = [HNO2(aq)].
⎝ [acid ] ⎠
As pKa for HNO2 = 3.15, [NO2-(aq)] = 1.00 M and [HNO2(aq)] = 0.075 M:
! [0.100] $
pH = 3.15 + log10 #
& = 3.27
" [0.075] %
pH = 3.27
If you wanted to adjust the pH of the mixture of Solution A
and Solution B to be exactly equal to 3.00, which component
in the mixture would you need to increase in concentration?
acid (HNO2)