Download Chemistry Practice Midterm: Titration Curves, Buffer Solutions, and Equilibria and more Exams Chemistry in PDF only on Docsity! 1 Chemistry 14BL Practice Midterm Key 1. Sketch the titration curves for each of the following systems. Calculate and indicate on each sketch, the equivalence point volume, the initial pH of the solution, and at least one other pH on the titration curve. Also be sure to indicate the approximate pH of the equivalence point if this is not the second point that you choose. (a) 10.00 mL of 0.0500 N HNO3 (beaker) titrated with 0.100 N NaOH (buret) This is the titration of a strong acid with a strong base. The shape will look like the data sets for the HCl –NaOH which are available on VOH. The equivalence point volume for the base is 5.00 mL. The initial pH is due to the complete dissociation of the nitic acid. [H+] = 0.05, pH = 1.3 The equivalence point pH is 7 (Calculation of other points require more arithmetic. Any other pH can be calculated prior to the equivalence point by considering the untitrated acid in the flask. For example at the 2.5 mL point in the titration, 5mL of acid are unreacted in a solution volume of 12.5 mL. [H+] = 0.020; pH = 1.7 After the equivalence point, you consider the concentration of the unreacted excess hydroxide.) (b) 10.00 mL of 0.0500 N benzoic acid (Ka = 6.46 x 10-5) (beaker) titrated with 0.100 N NaOH (buret). This is the titration of a weak acid with a strong base. The shape will look like your data from the lab when you titrated vinegar with NaOH. The equivalence point volume for the base is 5.00 mL. The initial pH is due to partial dissociation of thebenzoic acid. [H+] = 0.0018, pH = 2.7. Half way through the titration, after 5 mL of base have been added, the pH = pKa = 4.19) Because of the hydrolysis of the benzoate salt at the equivalence point, the pH will be greater than 7. 2. A buffer solution prepared by mixing NaHCO3 with K2CO3 has a pH of 10.6. (a) If the solution contains 0.133 M NaHCO3, what must the molarity of K2CO3 be? HCO3-(aq) = CO32- (aq) + H+(aq) Ka = 4.80 x 10-11 If pH = 10.6, [H+} = 2.51 x 10-11 The equilibrium expression for the reaction is [CO32-] [H+] = [CO32-] 2.51 x 10-11 = = 4.80 x 10-11 2 [HCO3-] 0.133 [CO32-] = 4.80 (0.133)/2.51 = 0.254 M (b) Explain how this system works as a buffer to offset large changes in pH. The system removes added acid by reacting with carobonate to give bicarbonate. Conversely bicarbonate reacts with added base to produce more carbonate and decrease the effect of the added base. (c) Carbon dioxide, which dissolves readily in water according to the following reactions, CO2 + H2O = H2CO3 H2CO3 = HCO3- + H+ accounts for the “fizz” in soda pop and affects its pH. Explain how the pH of 7-Up will change as it goes flat and CO2 bubbles from the solution. If you combine the two reactions you can more readily see the equilibrium between carbon dioxide bicarbonate CO2 + H2O = HCO3- + H+ [HCO3-][H+] = K [CO2] By Le Chatalier’s Principle, if CO2 is removed from the system the position of equilibrium will move to the left and you will decrease the hydrocen ion concentraiton in the system. The the pH of the system will increase as the soda goes flat. Alternatively, from the mass action equilibrium expresion you can see that if CO2 decreases so must the numerator in order for K to reamin constant. 3. Define the following terms being careful to identify any differences or relationships between them. (a) End point and equivalence point The end point in a titration is where the indicator changes color; the equivalence point is the volume when the same number of equivalents of acid and base have been mixed together. (b) Volumetric and non-volumetric glassware
