School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus
CHEM120R - 2013
Revision Tutorial – Equilibrium
1.
Calculate the pH at 25 oC of a 1.42  10–3 mol dm-3 solution of NaOH (aq).
[OH–] = 1.42  10–3 mol dm–3
pOH = 2.848
 pH = 14.00 – 2.848
= 11.15
2.
What is the concentration (in M) of hydronium ions in a solution at 25.0 °C with
pH = 4.282?
pH = - log[H3O+]
4.282 = -log[H3O+]
[H3O+] = antilog (-4.282)
= 5.22 105
3.
Calculate the concentration (in M) of hydronium ions in a solution at 25.0 °C with a pOH
of 4.223.
pOH = 4.223
pH = 14- 4.223 = 9.777
pH = - log[H3O+]
9.777 = - log[H3O+]
[H3O+] = antilog (-9.777)
= 1.67 1010
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4.
What is the pH of a 0.015 M aqueous solution of barium hydroxide?
Ba(OH)2  Ba2+ + 2OHBa(OH)2/1 = OH/2
Therefore [OH-]= 2[Ba(OH)2]
= 2  0.015 M
= 0.030 M
pOH = 1.52
pH = 14.00 - 1.52
= 12.48
5.
What is the pOH of a 0.020 M solution of barium hydroxide?
Ba(OH)2  Ba2+ + 2OHBa(OH)2/1 = OH/2
Therefore [OH-]= 2[Ba(OH)2]
= 2  0.020 M
= 0.040 M
pOH = 1.40
6.
What is the pOH of a 0.0384 M solution of HClO4?
pH = –log(0.0384) = 1.416.
pOH = 14.00 – 1.416 = 12.58
7.
Calculate the pH of a solution when 25. 00 mL of a 0.050 M HCl is mixed with 25.00 mL
of water.
Mi Vi = Mf Vf
Mf = Mi Vi /Vf
= 0.050 mol dm-3 x 0.025 dm3/0.050 dm3
= 0.025 mol dm-3
2
pH = -log [H+]
= - (-1.60)
8.
= - log (0.025 mol dm-3)
= 1.60
For the reaction
N2O4(g) ⇌ 2 NO2(g)
the concentrations of the substances present in an equilibrium mixture at 25 oC are
[N2O4] = 4.27 x 10-2 mol L-1
[NO2] = 1.41 x 10-2 mol L-1
What is the value of Kc for this temperature?
9.
For the reaction
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
Kc is 0.771 at 750 oC. If 0.0100 mol of H2 and 0.0100 mol of CO2 are mixed in a 1 litre
container at 750 oC, what are the concentrations of all substances present at equilibrium?
Solution
H2(g) +
CO2(g)
⇌ H2O(g) + CO(g)
I(mol)
0.0100
0.0100 ⇌
0
0
C
-x
-x
+x
+x
______________________________________________________
E
0.0100—x
0.0100—x
+x
+x
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10.
An industrial chemist puts 1.00 mol each of H2(g) and CO2(g) in a 1.00 L container
o
800 C. Kc = 0.923 for the reaction:
H2(g) + CO2(g) ⇌ H2O (g) + CO(g)
Calculate
(a)
the equilibrium concentrations in mol L-1.
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
1.00
-x
+ 1.00
-x
1.00-x
1.00-x
0
+x
0
+x
+x
+x
Kc = [H2O][ CO]/[ CO2][ H2(g)]
0.923 = x2/(1-x)2
x/1-x = 0.961
solve for x
= 0.500 M = [H2O] = [CO]
[H2] = [CO2] = 1.00 – x = 0.500 M
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at
(b)
Is the reaction at equilibrium if the [H2] = 0.25 M, [CO2] = 0.50 M, [CO] = 0.75 M and
[H2O] = 1.0 M. If not in which direction will the equilibrium shift? Show the appropriate
calculations.
Qc = [H2O][ CO]/[ CO2][ H2(g)]
= (0.750)(1.0)/ (0.25)(0.5)
= 6.0
Qp > Kc i.e 6.0 > 0.923
Reverse rxn will occur until Qc = Kc
11.
Formic acid, HCO2H, is a monoprotic acid with Ka = 1.82 x 10-4. Calculate the pH of a
0.153 M solution of formic acid.
⇌
HCO2-(aq)
HCO2H(aq)
Initial [ ]
0.153
0
0
-x
+x
+x
0.153 – x
+x
+x
Change in [ ]
Equilibrium [ ]
[HCO2-(aq)][ H+(aq)]
Ka =
= 1.82 x 10-4
[HCO2H(aq)]
Ka =
(x)(x)
= 1.82 x 10-4
0.153 - x
Assume that 0.153 >>> x , such that 0.153 – x = 0.153
x2 = 0.153 x 1.82 x 10-4 = 2.78 x 10-5
x = 5.28 x 10-3 M = [ H+(aq)]
Check assumption: (5.28 x 10-3/0.153) x 100% = 3.45%  valid
pH = - log [H+] = - log (5.28 x 10-3) = 2.278 = 2.28
5
+
H+(aq)
Reaction
12.
A solution of hydrazine, N2H4, has a concentration of 0.35 mol dm-3. Determine the pH of
the solution, and the percentage ionization of hydrazine? Hydrazine has Kb = 1.7 x 10-6.
N2H4 + H2O ⇌ N2H5+ + OH¯
Initial Conc.
Change
Equil. Conc.
N2H4
0.35
-x
0.35 - x
H2O
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N2H5+
0
+x
x
OH¯
0
+x
x
[N 2 H 5 ][OH  ]
Kb 
[N 2 H 4 ]
x2
 1.7  10 6
(0.35  x)
Assume x is small compared to 0.35 :
x2
 1.7  10 6
0.35
x  7.7  10 4  [OH  ]
pOH  log[OH  ]
 -log(7.7  10 4 )
 3.11
pH  14.00  pOH
 14.00  3.11
 10.89
7.7  10 4 mol dm 3
percent ionisation 
 100%
0.35mol dm 3
 0.22%
0.22%  5%
therefore assumption is valid .
13.
An 0.0284 M aqueous solution of lactic acid is found to be 6.7% ionized.
(i)
Determine Ka for lactic acid.
HC3H5O3(aq) + H2O (aq) ⇌ C3H5O3-(aq) + H3O+(aq)
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Percent Ionization = [H+]eq/ 0.0284 M  100
= 6.7%
[H+] = 0.0019 M
Ka = [H+] [C3H5O3-]/[ HC3H5O3]
= (0.0019 M)2/0.0284 – 0.0019
= 1.4 x 10-4
ii)
What is the pOH of the solution in question 13(i) above.
pH = −log [H3O+] = -log (0.0019) = 2.70
Therefore pOH = 14.00 – 2.70 = 11.30
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