One of the major problems is the large amount of phthalic acid (see your text fo

Question

One of the major problems is the large amount of phthalic acid (see your text for Kas) managed to make its way into the pond. The Company wants to know the pH of the pond. Cornelius makes some quick measurements and finds the pond is perfectly circular and has a radius of 1.000 meter and a depth of 1.000 meter. (Remember that 1 mL = 1 cm3). The tank that held the phthalic acid was a 3.00 L tank. The concentration in the tank was 3.000M.

A)What is the formal concentration, (H2A)0, in the pond if all of the phthalic acid solution is added?

B)What would the pH of the pond be, assuming there are no other acid/base species present except the phthalic acid?

C)What would the fraction of the three different acid species (H2A, HA- and A2-) be in solution at that pH?  

Explanation / Answer

1)

Now initial concentration of acid in tank = 3M

Volume = 3L

It is splitted into pond with volume = area X length = 3.14 x r x r X l = 3.14 m^3 = 3140 L

So new concentration will be

M1 V1 = M2 V2

3 X 3 = M2 X 3140

M2 = 0.0028 M

a) Ka of phthalic acid is 1.3 X 10^-3

As ka2 is very low so we will ignore it

pH = -log[H+]

phthalic acid ---> phthalic anion + H+

So Ka = [H+] [ phthalic anion] / [phthalic acid]

1.3 X 10^-3 = x^2 / (a-x)

We can ignore x in denominator

1.3 X 10^-3 = x^2 / 0.0028

x^2 = 3.64 X 10^-6

x = 1.907 X 10^-3

So [H+] = 1.907 X 10^-3

pH = 2.72

c) the [H+] = [HA-]

now let us consider the dissociation of HA- to A-2 and H+

Ka2 = 3.1 X 10^-6

Ka2 = [H+] [A-2] / [HA-]

3.1 x 10^-6 = x^2 / 1.907 X 10^-3

x^2 = 5.91 X 10^-9

x = 7.68 X 10^-5

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