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The graph below represents the titration of 50 mL of a 0.10 M acid with 0.10 M NaOH 024681012140102030405060V base added (mL)pHWhich of the following statements are consistent with this titration curve? X. The equivalence point occurs when approximately 25 mL of NaOH are added Y. The curve represents the titration of a weak acid. Z. The pKaof the acid is approximately 9.4. A. X,Y, and Z B. X,Y only C. Y, Z only D. Y only E. Z only X. False. The equivalence point occurs when exactly 50 mL of NaOH are added. Y. True. By the fact that the equivalence point is approximately 9 is consistent with a strong base titrating a weak acid. In addition, by the fact that the initial point is approximately 3 is consistent with the solution in the flask being a weak acid. Z. False. Henderson-Hasselbalch equation: pH = pKa+ log([A-]/[HA]) [A-] = [HA] at the half-way point, which is 25 mL of titrant. At the half-way point log([A-]/[HA]) = 0. Therefore, pH = pKa. Since at 25 mL of titrant (the half-way point) thepH = 6.0, then pKa= 6.0.
3844 Chem 162-2007 Final exam + answers Chapter 15B – Applications of Acid & Base Equilibria Titrations and indicators Calculate the pH after 10.0 mL of 0.400 M NaOH is added to 20.0 mL of 0.50 M CH3COOH. (Ka CH3COOH = 1.8 × 10í5). A. 2.52 B. 0.70 C. 13.70 D. 4.57 E. 7.00 (0.400M NaOH x 0.0100L)/(0.0100L + 0.0200L) = 0.1333 M NaOH (0.500M HA x 0.020L)/(0.020L + 0.010L) = 0.3333M HA HA + OH-moH2O + A-Reactions of strong bases go to completion. HA + OH-moH2O + A-Initial 0.3333 0.1333 0 Change -0.1333 -0.1333 +0.1333 Equilibrium 0.2000 0 +0.1333 Henderson-Hasselbalch equation: pH = pKa+ log([A-]/[HA]) pH = -log(1.8 x 10-5) + log((0.1333)/(0.2000)) = 4.57 CHEM 162-2007 EXAM II + ANSWERS CHAPTER 15B – APPLICATIONS OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) TITRATIONS AND INDICATORS 24. A titration curve is shown below. In this curve, the “titrant” refers to the substance in the buret, which is added to another substance in the titrating flask.