Midterm #2.2 Ans

Midterm #2.2 Ans - Chemistry 20L Final Exam Study Questions...

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Chemistry 20L Final Exam Study Questions Key Winter 2007 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 0.0.0500 M HNO 3 (beaker) titrated with 0.100 M NaOH (buret) The titration of a strong acid (HNO 3 ) with a strong base (NaOH) yields a titration curve with a characteristic slow-rising, flat portion followed by a steep rise that levels off once again. 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 + ] = 0.0500 M, pH = 1.30 The equivalence point pH is 7 Calculation of other points requires 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, 5 mL of acid are unreacted in a solution volume of 12.5 mL. [H + ] = 0.0200; pH = 1.70 After the equivalence point, you consider the concentration of the unreacted excess hydroxide. (b) 10.00 mL of 0.0500 M benzoic acid (K a = 6.46 x 10 -5 ) (beaker) titrated with 0.100 M NaOH (buret). The titration of a weak acid (benzoic acid) with a strong base (NaOH) yields a titration curve with a shape that resemebles your titrations 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 the benzoic acid. [H + ] = 0.00180 M, pH = 2.75 Half way through the titration, after 2.5 mL of base have been added, the pH = pK a = 4.19. Because of the hydrolysis of the benzoate salt at the equivalence point, the pH will be greater than 7. Calculation of [OH - ] at the equivalence point gives 5.75 x 10 -6 M or pOH = 5.2 and pH = 8.8
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2. A Chemistry 20L student determined the sodium tripolyphosphate (Na 5 P 3 O 10 ) in a detergent by ashing a sample and then titrating the acid in the ash with NaOH assuming the following reaction: OH - + H 2 PO 4 - "# HPO 4 2- + H 2 O (a) Based on the following data, calculate the percent STTP in the detergent. Molecular weight of STTP
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Midterm #2.2 Ans - Chemistry 20L Final Exam Study Questions...

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