Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Consider the titration of a 21.0-mL sample of 0.110 M HC2H3O3 (Ka=1.8×10−5) with 0.120 M NaOH...
1. Determine the initial pH and the volume of added base required to reach the equivalence point. 2. Determine the pH at 5.00 mL of the added base, the pH at one-half of the equivalence point, and the pH after adding 6.00 mL of base beyond the equivalence point..
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- Consider Titration of 25 mL of .1 M CH3CO2H With .1 M NaOH. Calculate The pH After Adding The Following Volumes of NaOH: a. 0 mL b. 12.5 mL. c. 20 mL d. 25 mL. e. 37.5 mL. I do know the answers, I just don't know the way to get to these answers. a. pH = 2.87 b. pH = 4.74 c. pH = 5.34 d. pH = 8.72 e. pOH = -log(.0192 M) = 1.72 pOH 14-1.72 = 12.28 pH (I just know the last steps to the answer for part e, but don't know all the other steps to get to answer).arrow_forwardFind the pH of the reaction mixture after the addition of the given volume (see below for your assignment) of 0.25M hydrochloric acid to a 20.0 mL sample of .15M potassium fluoride (Ka for HF =7.2 x 10^-4) 13.5 mL HCl added Weak base, strong acid pH curve Titration, post equivalence pointarrow_forwardConsider Titration of 25 mL of .1 M CH3CO2H With .1 M NaOH. Calculate The pH After Adding The Following Volumes of NaOH: a. 0 mL b. 12.5 mL c. 20 mL d. 25 mL e. 37.5 mL I do know the correct answers, I just don't how to get to the answers. a. pH = 2.87 b. pH = 4.74 c. pH = 5.34 d. pH = 8.72 e. pOH = -log (0.0192 M) = 1.72 14 - 1.72 = 12.28 (I been having a hard time people getting the correct answer for part e, people keep getting 12.30 as the answer, but it is not. The correct answer is 12.28. I only remember how to do the last steps to get part e answer, which is 12.28, as it can be seen).arrow_forward
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