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.

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.

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?

Thanks for clearing that up, I try to work all problems w/o a calculator and I'm continually frustrated by some of the Kaplan questions/explanations which expect you to use one.