Chemistry

Phthalic acid, H2C8H404, is a diprotic acid used in the synthesis of phenolphthalein indicator. Ka1= 1.2x 10^-3, and Ka2 = 3.9 x 10^-6. a) Calculate the hydronium-ion concentration of a .015 M solution. b) What is the concentration of the C8H4O4^-2 ion in the solution?

I don't know where to start!

To simplify things, let's call phthalic acid H2P.

Then H2P ==> H^+ + HP^- and
HP^- ==> H^+ + P^=

k1 and k2 are separated enough that the pH will be determined, ALMOST completely by k1. So think in terms of a simple acid, such as acetic acid; i.e.,
(H^+) = (HP^-)
and k1 = (H^+)(HP^-)/(H2P)
If this hint isn't sufficient, tell me what you are having trouble with and we can go from there.

Okay, that makes sense! I tried to calculate x2/.015=1.2x10^-3, but I got 4.24x10-3. The book says the answer is 3.7x10-3. What am I doing wrong?

If we call (H^+)= y, then
(HP^-) = y
and (H2P) = 0.015-y

Ka = (H^+)(HP^-)/(H2P)
Ka = (y)(y)/(0.015-y)=1.2E-3
If y in the denominator is neglected then your answer of 4.24E-3 is correct. However, y can't be neglected. If not neglected, then you must solve the resulting quadratic equation (I solved it by successive approximations because that is simpler than the quadratic formula). Anyway, I found 3.7E-3 which is what your book gave as the answer. You can see there is a significant difference between 4.2E-3(when 0.015-y = 0.015; that is, we neglect the y) and 3.7E-3(0.015-y where y is not neglected).

The second part of the problem is done this way.
Write the expression for k2.
k2 =(H^+)(P^=)/(HP^-)
From the problem you just worked for part a, we said (H^+)=y and (HP^-) = y; that is (H^+)=(HP^-)
Looking at the expression for k2, (H^+) in the numerator cancels with (HP^-) in the denominator and k2 = (P^=) or
(P^=) = 3.9E-6

Thank you so much! :)

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asked by Rachael
  1. Phthalic acid, H2(C8H404), is a diprotic acid used in the synthesis of phenolphthalein indicator. Ka1= 1.2x 10^-3, and Ka2 = 3.9 x 10^-6. a) Calculate the hydrogen-ion concentration of a .015 M solution. b) What is the concentration of the C8H4O4^2- ion in the solution?

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    posted by Crystal

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