Consider the titration of 60.0 mL of 0.100 M NH3 Kb= 1.8 x 10^-5 with 0.150...

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Consider the titration of 60.0 mL of 0.100 M NH3 Kb= 1.8 x 10^-5with 0.150 M HCl. Calculate the pH after the following volumes oftitrant have been added: 40.5 mL & 60.0 mL

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4.5 Ratings (865 Votes)

mmoles of NH3 = 60.0 x 0.100 = 6.00

Kb = 1.8 x 10^-5

pKb = 4.74 , pKa = 9.26

mmoles of HCl = 40.5 x 0.150 = 6.075

NH3 +Â Â HClÂ Â Â ------------>Â Â NH4+Cl-

6.00Â Â Â Â Â Â 6.075Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

0Â Â Â Â Â Â Â Â Â Â 0.075Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6.00

here strong acid remains. so

concentration of H+ = 0.075 / 60 + 40.5

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 7.46 x 10^-4 M

pH = - log (7.46 x 10^-4)

pH = 3.13

mmoles of HCl = 60 x 0.150 = 9

NH3 +Â Â HClÂ Â Â ------------>Â Â NH4+Cl-

6.00Â Â Â Â Â Â 9.00 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

0Â Â Â Â Â Â Â Â Â 3.00 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6.00

here strong acid remains. so

[H+] = 3 / 120 = 0.025 M

pH = 1.60

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Consider the titration of 60.0 mL of 0.100 M NH3 Kb= 1.8 x 10^-5 with 0.150...

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60.2K

Verified Solution

Question

Chemistry

Consider the titration of 60.0 mL of 0.100 M NH3 Kb= 1.8 x 10^-5with 0.150 M HCl. Calculate the pH after the following volumes oftitrant have been added: 40.5 mL & 60.0 mL

Answer & Explanation Solved by verified expert

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