Question

In: Chemistry

Use the concentration and pH value from solution 9 and the Kw value to approximate the...

Use the concentration and pH value from solution 9 and the Kw value to approximate the Ka for acetic acid:

01. M Acetic acid(HC2H3O2) & pH = 2.30

Ka = 3.65

soln 9 is: 0.1 M NaC2H3O2

pH soln 9 is : 7.65

Expert Solution

we know that

Kw= Ka X Kb

For dissociation of sodium acetate

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

Initial 0.1 0 0

Change -x +x +x

Equilibrium 0.1-x x x

Kb = [OH-][CH3COOH] / [CH3COO-] = x² / 0.1-x

Given that pH = 7.65 therefore pOH = 14-7.65 = 6.35

[OH-]= antilog(-pOH)

[OH-]= 4.47 X 10^-7 = x

We can ignore x in denominator

Kb = (4.47 X 10^-7)² / 0.1

Kb = 1.99 X 10^-12

Ka = Kw / Kb = 5.025 X 10^-3

queen_honey_blossom answered 1 year ago

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