
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration of 42.2 mL of 0.0520 M CH3COOH.
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- Consider the titration of a 40.0 mL of 0.211 M weak acid HA (Ka = 2.7 x 10⁻⁸) with 0.100 M LiOH. c. How many mL of the LiOH would be required to reach the halfway point of the titration? d. What is the pH of the solution at the equivalence point? e. What would be the pH of the solution after that addition of 100.0 mL of LiOH?arrow_forwardDetermine the pH at the point in the titration of 40.0 mL of 0.200 M HNNH2 with 0.100 M HNO: after 10.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0 x 10-6. PREV 1 2 4 NEXT > Based on your ICE table and definition of Kb, set up the expression for Kb in order to determine the unknown. Do not combine or simplify terms. = 3.0 x 10-6 RESET [0] [0.200] [0.0200] [0.100] [0.140] [0.175] [x] [2x] [0.200 + x] [0.200 - x] [0.0200 + x] [0.0200 - x] [0.100 + x] [0.100 - x] [0.140 + x] [0.140 - x] [0.175 + x] [0.175 - x]arrow_forwardCalculate the pH for each of the cases in the titration of 25.0 mL of 0.110 M pyridine, C₁H₁N(aq) with 0.110 M HBr(aq). The Kb of pyridine is 1.7 x 10-⁹. before addition of any HBr pH = after addition of 12.5 mL of HBr after addition of 16.0 mL of HBr after addition of 25.0 mL of HBr after addition of 35.0 mL of HBr 9.31 pH = pH = pH = pH = 5.23 4.63 Incorrect 11.65 Incorrect 1.33 Incorrectarrow_forward
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