25.00 mL of a 0.1000 molar methylamine ( Kb = 4.4 x 10^-4 ) solution is placed i

ID: 853665 • Letter: 2

Question

25.00 mL of a 0.1000 molar methylamine ( Kb = 4.4 x 10^-4 ) solution is placed in a beaker to be titrated against a 0.1000 m HCl solution.

1. Find the pH of the methylamine solution before any HCl is added.

2. Find the pH of the solution in the beaker after 10.00 mL of HCl was added.

3. Find the pH of the solution in the beaker after a total of 25.00 mL of HCl was added.

4. Find the pH of the solution in the beaker after a total of 25.20 mL of HCl was added.

PLEASE SHOW ALL WORK AND EXPLAIN! Thank you!

1. pOH = 1/2[pKb+LOg C]
pKb = -log Kb = 4-log 4.4 =3.3565

=1/2[3.3565 + log0.1*25/1000]
= 0.378
pH = 14-0.378 = 13.622

2. pOH = pKb+LOg [10*0.1/35/0.1*25/35]
= 3.3565 ?0.39794
= 2.95856
pH = 14-2.95856 = 11.04

3. pOH = pKb+LOg [25*0.1/35/0.1*25/35]
= 3.3565 + 0
= 3.3565
pH = 14-3.3565 = 10.64

4.
pH      = 7-1/2[pKb+log C]
   = 7-1/2[3.3565+log 25.2*0.1/1000]
   = 6.621

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