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pH. Vol (mL). Midpoint 1. 2.99 Midpoint 1= w-2.99 Midpoint 2= w-6.56 Mass of unknown sample =.122g. Molarity of NaOH sample =.1 M. Equivalence point Midpoint 2. 1. 10.21 Molar ratio at equivalence point 1=1:1. Mol of acid → (.1) (.01021) =.001021122/.001021=119.5 /mol. 6.56 Equivalence point 2. 21.71

pH. Vol (mL). Midpoint 1. 2.99 Midpoint 1= w-2.99 Midpoint 2= w-6.56 Mass of unknown sample =.122g. Molarity of NaOH sample =.1 M. Equivalence point Midpoint 2. 1. 10.21 Molar ratio at equivalence point 1=1:1. Mol of acid → (.1) (.01021) =.001021122/.001021=119.5 /mol. 6.56 Equivalence point 2. 21.71

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can you find Ka1 and Ka2?

pH. Vol (mL). Midpoint 1. 2.99 Equivalence point 1. 10.21 Midpoint 1= w-2.99 Midpoint 2= w-6.56 Mass of unknown sample =.122g. Molarity of NaOH sample =.1 M. Molar ratio at equivalence point 1 = 1:1. Mol of acid → (.1) (.01021) =.001021122/.001021-119.5 5/mal. Midpoint 2. 6.56 Equivalence point 2. 21.71 Include an annotated calculation showing how you determined the Ka1 and Ka2 (i.e. How did you find Kǝ1 and Ka2?).
Transcribed Image Text:pH. Vol (mL). Midpoint 1. 2.99 Equivalence point 1. 10.21 Midpoint 1= w-2.99 Midpoint 2= w-6.56 Mass of unknown sample =.122g. Molarity of NaOH sample =.1 M. Molar ratio at equivalence point 1 = 1:1. Mol of acid → (.1) (.01021) =.001021122/.001021-119.5 5/mal. Midpoint 2. 6.56 Equivalence point 2. 21.71 Include an annotated calculation showing how you determined the Ka1 and Ka2 (i.e. How did you find Kǝ1 and Ka2?).
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