K_ = [CH_ COO^- H^+]/[CH_ COOH] = (0.100 + x)/0.100 - x = 1.75 times 10^-5 right

Question

K_ = [CH_ COO^- H^+]/[CH_ COOH] = (0.100 + x)/0.100 - x = 1.75 times 10^-5 rightarrow x^2 + (0.100 + 1.75 times 10^)x - 1.75 times 10^= 0 x = -1.75 times 10^+ squareroot (1.75 times 10^-5)^- 4 times 1 times (-1.75 times 10^)/2 times 1 rightarrow x = 1.749 times 10^-5 alpha = x/F = 1.749 times 10^-5/0.1000 = 1.749 times 10^-4 Calculate the fraction of acetate ions that undergo hydrolysis in an aqueous solution with formal concentration of 0.1000 M. What value does this fraction takes, if acetic acid is added to the solution so that the acid contributes a formal concentration of 0.1000 M?

Explanation / Answer

Fraction of acetate ions (A-) that hydrolyse when

[A-] = 0.1 M

a)

A- + H2O <-> HA + OH-

Kb = [HA][OH-]/[A-]

Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.55*10^-10

[HA] = x

[OH-] = x

[A-] = 0.1-x

Kb = [HA][OH-]/[A-]

5.55*10^-10 = x*x/(0.1-x)

x = OH- = 7.42*10^-6

then

[HA] = x= 7.42*10^-6

the fraction --> [HA-]hdyrolysed / [A-] not hydrolysed = (7.42*10^-6) / 0.1 = 7.42*10^-5 is hydrolysed

B)

if we add 0.1 Acetic acid:

pH = pKa + log(A-/HA)

pH = 4..75 + log(0.1/0.1)

pH = 4.75

note that

A- = HA

so, there is 50% hydrolysis

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