Question

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Cinnamic acid is obtained from oil of cinnamon. 125 millimoles of HCI were added to a solution containing 0.2 moles of a cinnamic acid and 0.3 moles of its conjugate base. After mixing, the pH of the solution was found to be 4.17. What is the pK, of cinnamic acid? A) 4.44 B) 4.17 C) 3.90 D) 4.93 E) 4.66 The digestive enzyme pepsin is secreted by stomach cells into an environment with a pH of 1.5-2. It has as a pH optimum of about 2.0 and is inactive at pH 6.5. What would be the buffer of choice to study this enzyme? 12. A) carbonic acid (Kal-4.5 x 10". Ка,-4.7 x 101') B) lactic acid (Ka = 1.38 x 10-4) C) tris-hydroxymethyl aminomethane (K -8.32 x 10) D) phosphoric acid (Kal.1 Ka 63 x 10: Ka3 E) boric acid (K,-5.75x 10'10) 4.5 x 10)

Answer 1

Concentration of cinnamic acid HX, [HX] = 0.2 mol Concentration of conjugate base, [X^-] = 0.3 mol pK_a of a cinnamic acid HX = 4.17 no of moles of acid added = 125 mm ol = 0.125 mol ICE table: X^- + H^+ rightarrow HX I(mol): 0.3 0.125 0.2 C(mol): - 0.125 - 0.125 + 0.125 E(mol): 0.175 0 0.325 Concentration of acid HX, [HX] = 0.325 mol Concentration of conjugate base, [X^-] = 0.175 mol According to Henderson equation, p^H = Pk_a + log [X^-]/[HX] = 4.17 + log 0.175 mol/0.325 mol = 3.90