Calculate the pH and fraction of dissociation () for each of the acetic acid (CH

Question

Calculate the pH and fraction of dissociation () for each of the acetic acid (CH3COOH, pKa = 4.756) solutions below. (a) A 0.00369 M solution of CH3COOH. (b) A 2.03 × 10-12 M solution of CH3COOH.

Hint: The pH of a solution is calculated from the concentration of H ions at equilibrium. For a solution of acetic acid, a weak acid, there are two sources that contribute to the H concentration. The first source is acetic acid. A generic weak acid (HA) dissociates in water into its conjugate base (A–) and H ions and is described by the acid dissociation constant, Ka.The second source of H ions is the dissociation of water, which is described by the dissociation constant Kw. In each of the acetic acid solutions, the contribution from one of these sources is negligible and can be ignored when calculating the H concentration.

Explanation / Answer

a)

we know that

pKa = -log Ka

so

4.756 = - log Ka

Ka = 1.754 x 10-5

we know that

for weak acids

[H+] = sqrt ( Ka x C)

given

Ka =

C = 0.00369

so

[H+] = sqrt ( 1.754 x 10-5 x 0.00369)

[H+] = 2.544 x 10-4

now

pH = -log [H+]

so

pH = -log 2.544 x 10-4

pH = 3.5944868

now

fraction of dissociation = [H+] / C

= 2.544 x 10-4 / 0.00369

= 0.068942

so

the fraction of dissociation is 0.068942

b)

now

[H+] = sqrt ( 1.754 x 10-5 x 2.03 x 10-12)

[H+] = 5.967 x 10-9

now

we know that

water has 10-7 M [H+]

so

total [H+] = 10-7 + 5.967 x 10-9

[H+] = 1.05967 x 10-7

so

pH = -log 1.05967 x 10-7

pH = 6.9748

so

the pH is 6.9748

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