2. 60.0 mL of 0.100 M NH, (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 = 6mmo NH3+ HCI NH+CI KA=! 0.15 x 10-15 mmol I 6 C 15 1.5 4.5mm 4.5mmol [N] 70mL kw kb = 110-H =5.56 PH = pKa + log (bass) pKa = log (S56x107) *109 1.8x10-3 =0.0643 pka = 9.26 [NH] 1.5mmol 70ML = 0.0214 PH-9.26+109 60.0643) PH=9.747 For the titration in question 2, calculate the pH at the equivalence point.

2. 60.0 mL of 0.100 M NH, (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 = 6mmo NH3+ HCI NH+CI KA=! 0.15 x 10-15 mmol I 6 C 15 1.5 4.5mm 4.5mmol [N] 70mL kw kb = 110-H =5.56 PH = pKa + log (bass) pKa = log (S56x107) *109 1.8x10-3 =0.0643 pka = 9.26 [NH] 1.5mmol 70ML = 0.0214 PH-9.26+109 60.0643) PH=9.747 For the titration in question 2, calculate the pH at the equivalence point.

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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