7.44. The ionization constant of phenol is 1.0 x 10-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?
7.44. The ionization constant of phenol is 1.0 x 10-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?
C6H5OH? C6H5O- + H+
| C6H5OH | C6H5O- | H+ |
Initial | 0.05 M | 0 | 0 |
After dissociation | 0.05 –x | x | x |
Ka = x2 / (0.05 - x) = 1.0 x 10-10
=> x2 / 0.05 = 1.0 x 10-10
=> x2 = 5 x 10-12
=> x= 2.2 x 10-6 M
In presence of 0.01 C6H5Na, suppose y is the amount of phenol dissociated, then at equilibrium
[C6H5OH] = 0.05 – y ≈ 0.05,
[C6H5O-] = 0.01 + y ≈ 0.01 M, [H+] =
Similar Questions for you
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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