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Calculate the pH of 0.10 M sodium acetate solution ( Ka, acetic acid = 1.76 x...

Calculate the pH of 0.10 M sodium acetate solution ( Ka, acetic acid = 1.76 x 10-5 ).

Solutions

Expert Solution

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/1.8*10^-5

Kb = 5.556*10^-10

CH3COO- dissociates as

CH3COO- + H2O -----> CH3COOH + OH-

0.1 0 0

0.1-x x x

Kb = [CH3COOH][OH-]/[CH3COO-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.556*10^-10)*0.1) = 7.454*10^-6

since c is much greater than x, our assumption is correct

so, x = 7.454*10^-6 M

use:

pOH = -log [OH-]

= -log (7.454*10^-6)

= 5.1276

use:

PH = 14 - pOH

= 14 - 5.1276

= 8.8724

Answer: 8.87

queen_honey_blossom answered 2 years ago
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