a. Calculate the pH of a 7.71×10^{-3} M solution of H_{2}SO_{4}.

(K_{a} = 0.0120 for HSO_{4}^{-})

b. Calculate [OH^{-}] of a 5.45×10^{-1} M aqueous solution of pyridine (C_{5}H_{5}N, K_{b} = 1.7×10^{-9}).

c. Calculate the pH of the above solution.

Question a)

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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