# Calculating pH from pKa

Discussion in 'MCAT Study Question Q&A' started by nfg05, Jun 18, 2008.

1. ### nfg05 2+ Year Member

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So I got the following question correct, but my method is quite different than the Kaplan explanation, which frankly I don't really follow. Here's the question and how I solved it, would really appreciate if someone could confirm my method is valid so I know I didn't just get it right by blind luck. I'd also like to hear easier/other ways of thinking about the question since these types always give me trouble.

If the pKa1 value for H2CO3 is 6.4, what is the pH of a 10^-3 M solution of this acid?

My solution:
pKa = -log(Ka) = 6.4
log(Ka) = -6.4
10^-6.4 = Ka = [H+][HCO3-]/[H2CO3]

Now the H2CO3, being a weak acid, will dissociate to give some concentration x of H+ ions and that same concentration x of HCO3- ions

Ka = (x)(x)/(10^-3 - x) = 10^-6.4

Ignoring the x in the denominator since it is small relative to 10^-3, we have:

x^2 = (10^-3)(10^-6.4) = 10^-9.4.
x = (10^-9.4)^(1/2) = 10^-4.7

Since x = [H+], pH = -log(x) = -log(10^-4.7) = 4.7

And pH = 4.7 is the correct answer given by Kaplan.

Kaplan starts off by saying " since the negative log of the dissociation constant is 6.4, the antilog is 4 x 10^-7 " (How the F do you get that exactly w/o a calculator?). But as per above, I got Ka = 10^-6.4 and got the correct answer. WTF...

I can post the rest of the long, rather confusing Kaplan explanation if people want to hear it, but I'd rather hear a straightforward approach other people have used successfully for these kinds of problems.

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Last edited: Jun 18, 2008
2. ### cwfergus 10+ Year Member

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Ive always done it your way, and will always do it that way haha

3. ### minhaj Awesomeness Incarnate 7+ Year Member

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what you did is in fact what kaplan but with a calculator.10^-6.4 = 3.98 x 10^-7. so your method is fine.

4. ### 161927 Guest

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your method seems faster than using an ICE (Initial Change Equilibrium) table

Does your method work for situations where there is not 1 : 1 stoichiometry?

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