DrBob has given you the equation to use:

k2 = [H^+][SO4^-]/[HSO4^-]

and the values to substitute

[H^+] = 0.01 + x
[SO4^-2] = x
[HSO4^-] = 0.01 - x
k2 = 1.2 x 10^-2

you need to substitute these and solve for x.

Remember that you are after 0.01+x [H^+]
to calculate the pH.

is the answer ph of 2.398

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  1. I don't obtain that for an answer.

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  2. Nor do I. The first ionization of concentration .01 will give a pH of 2, so adding more H from the second ionization has to make the pH more acid (ie less than 2).

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  3. In one of your earlier posts, you asked if this was the correct quadratic equation and it was.
    X^2 + 0.022X - 1.2E-4 = 0
    Using the quadratic formula, which is
    X=(-b +/- sqrt[b^2-4ac])/2a
    solve for X. What is that value?

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