3. 2. 60.0 mL of 0.100 M NH3 (K=1.8 X 10-5) is titrated with 0.150 M HCI. Calculate the pH after the addition of 10.0 mL acid. 0.1x60m² = (mmol NH3+ HCI NH4+C5 Ka= 0.15 x 10me=15 mmol kw Kb (brize) 4.5mmol 1.8x10-3 = 556 of H = pKa + log (benze =0.0643 I 6 C 15 15 рка Eg Zom pka = 9.26 pKa = log (S5bx107) [MH] 1 7/0 mL = 0.0214 PH-9.26+109 (0.06+13) PH=9.74 For question 2, at the given volume for acid addition the titration is considered to be in the "buffer region". Explain why. 6.0214

3. 2. 60.0 mL of 0.100 M NH3 (K=1.8 X 10-5) is titrated with 0.150 M HCI. Calculate the pH after the addition of 10.0 mL acid. 0.1x60m² = (mmol NH3+ HCI NH4+C5 Ka= 0.15 x 10me=15 mmol kw Kb (brize) 4.5mmol 1.8x10-3 = 556 of H = pKa + log (benze =0.0643 I 6 C 15 15 рка Eg Zom pka = 9.26 pKa = log (S5bx107) [MH] 1 7/0 mL = 0.0214 PH-9.26+109 (0.06+13) PH=9.74 For question 2, at the given volume for acid addition the titration is considered to be in the "buffer region". Explain why. 6.0214

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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