Question

In: Chemistry

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

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

Expert Solution

mmoles of NH3 = 60.0 x 0.100 = 6.00

Kb = 1.8 x 10^-5

pKb = 4.74 , pKa = 9.26

a)

mmoles of HCl = 40.5 x 0.150 = 6.075

NH3 + HCl ------------> NH4+Cl-

6.00 6.075 0

0 0.075 6.00

here strong acid remains. so

concentration of H+ = 0.075 / 60 + 40.5

= 7.46 x 10^-4 M

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

pH = 3.13

b)

mmoles of HCl = 60 x 0.150 = 9

NH3 + HCl ------------> NH4+Cl-

6.00 9.00 0

0 3.00 6.00

here strong acid remains. so

[H+] = 3 / 120 = 0.025 M

pH = 1.60

queen_honey_blossom answered 1 year ago

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