Question

A 200 mL solution of 0.100 M solution of Ammonia, NH3 , is slowly mixed with 0.100 M HCl. Determine the pH after addition of:

0 mL of HCl, 50 mL of HCl, 100 mL of HCl, 200 mL of HCl, 250 mL of HCl

4. Describe in exact detail (including gram quantities and identity of species) how you would create a 500 mL solution of buffer with a pH 8.0 ?    

Answer 1

1)

millimoles of pyridine = 200 x 0.1 = 20

pKb = 4.74

a) 0.0 mL addition of HCl

pOH = 1/2 [pKb -logC] = 1/2 [4.74 -log0.1] = 2.87

pH + pOH = 14

pH = 11.13

b) after the addition of 50 mL HCl

millimoles of HCl = 50 x 0.1 = 5

NH3 + HCl   -------------> NH4Cl

20          5                           0

15         0                           5

pOH = pKb + log [salt / base]

        = 4.74 + log [5 / 15]

       = 4.26

pH = 9.74

b) after the addition of 100 mL HCl

it is hlalf equivalence point : so

pOH = pKb

pOH = 4.74

pH +pOH =14

pH = 9.26

d) after the addition of 200 mL HBr

it is equivalence point only salt is formed

salt millimoles = 20

salt concentration = millimoles / total volume = 20 / (200+200) = 0.05 M

salt is from strong acid weak base so pH <7

pH = 7 -1/2 [pKb + logC]

pH = 7 - 1/2 [4.74 + log 0.05]

pH = 5.28

e) after the addition of 250 mL HBr

millimoles of HCl = 250 x0.1 = 25

strong acid remained in the solution

[H+] =5 / (250+200) = 0.011 M

pH = -log(0.011)

pH = 1.95