Question

In: Chemistry

Calculate the pH of a 0.51 M CH3COOK solution. (Ka for acetic acid = 1.8×10−5.)

Calculate the pH of a 0.51 M CH₃COOK solution. (K_a for acetic acid = 1.8×10⁻⁵.)

Expert Solution

Kb = Kw/Ka

Kw is the 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-

0.51 0 0

0.51-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)*0.51) = 1.683*10^-5

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

so, x = 1.683*10^-5 M

use:

pOH = -log [OH-]

= -log (1.683*10^-5)

= 4.7739

use:

PH = 14 - pOH

= 14 - 4.7739

= 9.2261

Answer: 9.23

queen_honey_blossom answered 2 years ago

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