The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
1) CH3COOH ↔ CH3COO- + H+ Ka = 1.74x10-5
2) H2O + H2O ↔ H3O+ + OH- Kw = 1.0x10-14
Since Ka > Kw
CH3COOH + H2O ↔ CH3COO- + H3O+
Ci = 0.05 0 0
0.05- 0.05α 0.05α 0.05α
ka = 0.05α x 0.05α / 0.05- 0.05α
= 0.05α x 0.05α / 0.05 (1-α)
= 0.05α2 / (1-α)
⇒ 1.74x10-5 = 0.05α2 / (1-α)
⇒ 1.74x10-5 - 1.74x10-5 α = 0.05α2
⇒ 0.05α2 + 1.74x10-5 α - 1.74x10-5 =0
D = b2 - 4ac
= (1.74x10-5 )2 - 4 (0.05) (1.74x10-5)
= 3.02x10-25 + 0.348x10-5
Method 2:
Degree of Dissociation,
CH3COOH ↔ CH3COO- + H+
Thus, concentration of CH3COO- = c.α
= 0.05x1.86x10-2
= 0.93x10-2
=.00093M
Since [oAc-] = [H+]
[H+] = .00093 = 0.093x10-2
pH = -log[H+]
= -log (0.93x10-2)
∴ pH = 3.03
Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.
Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH
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a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH4 (g) + H2O (g) ↔ CO (g) + 3H2 (g)
(a) Write as expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst ?
At 700 K, equilibrium constant for the reaction:
H2 (g) + I2 (g) ↔ 2HI (g)
is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 ×10-3. If decomposition is depicted as,
PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ΔrH0 = 124.0 kJ mol–1
(a) write an expression for Kc for the reaction.
(b) what is the value of Kc for the reverse reaction at the same temperature ?
(c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is increased ?
Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl (g) ↔ 2NO (g) + Cl2 (g); Kp = 1.8 × 10–2 at 500 K
(ii) CaCO3 (s) ↔ CaO(s) + CO2(g); Kp = 167 at 1073 K
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2N2 (g) + O2 (g) ↔ 2N2O (g)
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10–37, determine the composition of equilibrium mixture.
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(i) 2NOCl (g) ↔ 2NO (g) + Cl2 (g)
(ii) 2Cu(NO3)2 (s) ↔ 2CuO (s) + 4NO2 (g) + O2 (g)
(iii) CH3COOC2H5(aq) + H2O(l) ↔ CH3COOH (aq) + C2H5OH (aq)
(iv) Fe3+ (aq) + 3OH– (aq) ↔ Fe(OH)3 (s)
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(i) CH2=C=O,
(ii) CH3CH=CH2,
(iii) (CH3)2CO,
(iv) CH2=CHCN,
(v) C6H6
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(iii) CH4
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(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12 H2O
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(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
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(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
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(i) normal water
(ii) acidified water, and
(iii) alkaline water?
Write equations wherever necessary.
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(ii) Lead(IV) chloride is highly unstable towards heat.
(iii) Lead is known not to form an iodide, PbI4.
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