What is the pH ( 3 significant digits) of a 0.01 M solution of sulfuric acid? Hint is that the first ionization constant is "large", and the second is finite, and the [H+] donated by the second equilibrium depends upon the amount of [H+] already donated by the first ionization. What is the answer please.
For Further Reading
chemistry - DrBob222, Wednesday, March 26, 2008 at 1:29am
H2SO4 ==> H^+ + HSO4^-
HSO4^- ==> H^+ + SO4^-2
The first ionization is complete; i.e., it produces 0.01 M H^+.
The second ionization is not complete and obeys k2 = (H^+)(SO4^-)/(HSO4^-)
The easiest way to work the problem, I think, is to write the expression for k2, look up k2 for its value, then substitute as follows:
(H^+) = 0.01 + x
(SO4^-2) = x
(HSO4^-) = 0.01 - x
Then solve the quadratic for x.
Total (H^+) then is 0.01 + x.
The k2 value is 1.2 x 10^-2.. where exactly am I substituting that? can you show me how using that number.
For Further Reading
chemistry - Dr Russ, Wednesday, March 26, 2008 at 9:20am
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.
I got the answer of either 4.0 or 1.65.. but I don't know if it is right.. Can you refresh my memory on how to solve for x in this equation please.
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