a. Calculate the pH of a 7.71×10-3 M solution of H2SO4.
(Ka = 0.0120 for HSO4-)
b. Calculate [OH-] of a 5.45×10-1 M aqueous solution of pyridine (C5H5N, Kb = 1.7×10-9).
c. Calculate the pH of the above solution.
Please you have to solve this equation:
0.012 = y^2 / (0.00771 - y)
where y is the concentration of [H+], please keep in mind that since H2SO4 is a strong acid, then your chemical reaction is:
H2SO4 = H+ + SO4-.
After a simple computation, I get:
y=[H+] = 0.0053= 5.3*10^(-3)
so we have_
pH = -Log(5.3*10^(-3))= 2.28
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