Question

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7. Ibuprofen is the active ingredient of the over-the-counter painkiller Advil. Its structure is shown below. Like other non-steroidal anti-inflammatory drugs, it works by inhibiting the synthesis of prostaglandins involved in the inflammation, pain, and fever pathway of the body. What is the Ka of ibuprofen if a 0.55 M aqueous solution of ibuprofen has a pH of 2.58 at 25°C? OH A. 1.45 x 10-3 B. 1.26 x 10-5 C. 7.31 x 10-7 D. 4.59 x 10-8 E. 2.63 x 10-3 8. A titration enthusiast titrates a 500 mL solution of 2 M C2H5OCOOH (lactic acid) with a 1M KOH solution for the sheer thrill of titration. What is the pH after 500 mL of the KOH has been added at 25°C? Structure of lactic acid: YOH ОН Ka of lactic acid = 1.4 x 10-4 A. 5.8 B. 10.1 C. 4.7 D. 3.9 E. 12.1 9. Which one of the following solutions will have the highest pH at 25°C? A. 1 M NaClO2 (aq) B. 1 M NaClO4 (aq) C. 1 M NaBrO2 (aq) D. 1 M NaBrO4 (aq) E. 1 M NaCl(aq)

Answer 1

Initial conc. of weak acid HA ibuprofen = 0.55 M The reaction of HA in water is Let x be the change in the conc. of each species at equilibrium. ka HA + H2O = H204 + A Initial 0.55 Change -X Equilibrium 0.55-X Given that pH = 2.58 [H+]= 10-2.58 = 0.00263 So, x = 0.00263 The acid dissociation constant is x.x Ka 0.131- x (0.00263)(0.00263) - 1.26x10-3 Ko 0.55-(0.00263) Option B is the right answer.

Number of moles of Lactic acid = M*V = 2 M * 500 mL = 1000 mmol Number of moles of KOH = 1M* 500 mL = 500 mmol Net moles of acid = 1000 - 500 = 500 mmol This is half equivalence point. (mid point) At this point, pH = pka = -log(1.4 x 10-4) = 3.9 Option D is the right answer. 3. Nacl04, NaBro4 and NaCl are the salts formed by the reaction of strong acids and strong base. So, pH of these solutions is 7 HClO2 and HBrO2 are weaker acids. HClO2 is stronger acid than HBrO2 HBrO2 is the weakest acid and its conjugate base Bro2- will be the strongest base. Hence NaBro2 will have the highest pH Option C is the right answer.

concentration of plamitic acid (C15H3COOH) = 0.050 M Ką of C15H3COOH = 1.66 x 10-5 Let x be the change in the conc. of each species at equilibrium. M C15H3COOH + H2O P C15H31C00 + Initial 0.050 Change Equillibrium 0.050 - x x.x ha0.050- X -=1.66x10-5 x is small, so, 0.050- x = 0.050 x.x = 8.30-10-7 3x=9.11x104 :: [H20*]=9.11x10+M pH =- logH,0*]=- log (9.11x104)= 3.040 = 3.0 pH = 3.0 Option D is the right answer.