7.59. The ionization constant of propanoic acid is 1.32 * 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
7.59. The ionization constant of propanoic acid is 1.32 * 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
Let the degree of ionization of propanoic acid be α. Then, representing propionic acid as HA, we have:
HA + H2O ⇔ H3O+ + A−
(.05−0.0α)≈0.5 0.05α 0.05α
Ka= [H3O+] [A−] / [HA]
= (0.05α) (0.05α) / 0.05
= 0.05α2
=> α = (Ka /0.05)1/2
=1.63*1
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0.01 M NaOH,
M = 1 * 10-2

pOH = 2
pH = 2
Kp = Kc (RT)Dng
36 * 10–2 = Kc (0.0821 * 300)–1
Kc = 0.36 * 0.0821 * 300 = 8.86 » 9
A(g) ->B(g) + (g)
Initial moles n 0 &nbs
On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.
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Chemistry Ncert Solutions Class 11th 2023
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