missy8005
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Can anyone help with this chemistry question number 5.
(a) The acid dissociation constant of phenol is 1.2x10¯¹° mol dm ¯³ . For a 0.01 mol dm ¯³ solution of the acid calculate
degree of dissociation

pH
b) repeat the calculation for the base dimethylamine kb = 7.4x10 ¯⁴
nb for this example you will require to solve a quadratic equation since kb is not small enough to assume(1-a)=1
Last edited by missy8005; 2 years ago
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username3249896
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Use Oostwald's dilution law for (a)
degree of ionisation = sqrt(1.2 x 10^-10 x 100) = 1.1 x 10^-4

[H+] = 1.1 x 10^-4 x 0.01 = 1.1 x 10^-6 so pH 5.96

for (b)
B + H2O -> BH+ + OH-
0.01, 0, 0
0.01-a, a, a

a^2/(0.01-a) = 7.4 x 10^-4
solve the quadratic a = 2.38 x 10^-3

degree of ionisation = 2.38 x 10^-3 / 0.01 = 0.238

[OH-] = 0.238 x 0.01 = 2.38 x 10^-3
Kw = 1 x 10^-14 so [H+] = 4.21 x 10^-12 M
hence pH = 11.38
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missy8005
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Many thanks

(Original post by BobbJo)
Use Oostwald's dilution law for (a)
degree of ionisation = sqrt(1.2 x 10^-10 x 100) = 1.1 x 10^-4

[H+] = 1.1 x 10^-4 x 0.01 = 1.1 x 10^-6 so pH 5.96

for (b)
B + H2O -> BH+ + OH-
0.01, 0, 0
0.01-a, a, a

a^2/(0.01-a) = 7.4 x 10^-4
solve the quadratic a = 2.38 x 10^-3

degree of ionisation = 2.38 x 10^-3 / 0.01 = 0.238

[OH-] = 0.238 x 0.01 = 2.38 x 10^-3
Kw = 1 x 10^-14 so [H+] = 4.21 x 10^-12 M
hence pH = 11.38
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