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.
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(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?
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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 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 ?
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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?
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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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(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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