a. Babituric acid, HC4H3N2O3, is used to prepare barbiturates, a class of drugs used as sedatives. A 325 mL aqueous solution of barbituric acid has a pH of 2.34 and contains 9 g of the acid. What is the Ka for barbituric acid?
b.Penicillin (MM = 356 g/mol), an antibiotic often used to treat bacterial infections, is a weak acid. Its Ka is 1.7 x 10^-3. Calculate [H+] in solutions prepared by adding enough water to following to make 725 mL.
Please show and explain work.
2007-03-29 16:37:06 · 1 answers · asked by nirmal 1 in Science & Mathematics ➔ Chemistry
a. MW barbituric acid = 128 g/mol, so 9 grams is 0.0703 moles. 0.0703 moles / 0.325 L = 0.216 M to start with. If pH = 2.34, then the [H+] = [C4H3N2O3-] = 10 ^ -2.34 = 0.00457 M. Ka = (0.00457)^2/(0.216-0.00457) = 9.88 * 10^-5.
b. In order to solve this, you need a weight of penicillin, because you need a molarity of penicillin to figure out the pH.
2007-03-29 16:45:22 · answer #1 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋