100 ml of 0.05 M ammonia NH3 was tittrated with 0.05 M sulfuric acid H2SO4. find

ID: 855476 • Letter: 1

Question

100 ml of 0.05 M ammonia NH3 was tittrated with 0.05 M sulfuric acid H2SO4. find out pH when :

- 0 ml of H2SO4 was added

- 10 ml of H2SO4 was added

- 20 ml of H2SO4 was added

- 30 ml of H2SO4 was added

- 40 ml of H2SO4 was added

- 45 ml of H2SO4 was added

- 49 ml of H2SO4 was added

- 49.5 ml of H2SO4 was added

- 50 ml of H2SO4 was added

- 50.5 ml of H2SO4 was added

- 51 ml of H2SO4 was added

- 55 ml of H2SO4 was added

- 60 ml of H2SO4 was added

~ Based on your answers, please draw a pH curve for this titration.

1) pOH = 1/2[pKa- log C]
   pKb of ammonia = 4.75
        = 1/2[4.75-log 0.5]
   = 2.526
pH = 14-pOh = 14-2.526 = 11.474

2) pOH= pKb + log[salt or Acid/Base]
pKb of ammonia = 4.75
pOH = 4.75 + log [ 0.05*1000/10/0.5]
= 5.75
pH = 14-pOH = 14-5.75 = 8.25

3) pOH = 4.75 + log [ 0.05*1000/20/0.5]
= 5.45
pH = 14-pOH = 14-5.45 = 8.55

4) pOH = 4.75 + log [0.05*1000/30/0.5]
= 5.27
pH = 14-pOH = 14-5.27 = 8.73

5) pOH = 4.75 + log [0.05*1000/40/0.5]
= 5.148
pH = 14-pOH = 14-5.148 = 8.852

6) pOH = 4.75 + log [0.05*1000/49.5/0.5]
= 5.055
pH = 14-pOH = 14-5.055 = 8.945

7) pOH = 4.75 + log [0.05*1000/50/0.5]
= 5.05
pH = 14-pOH = 14-5.05 = 8.95

8) pOH = 4.75 + log [0.05*1000/50.5/0.5]
= 5.047
pH = 14-pOH = 14-5.047 = 8.953

9) pOH = 4.75 + log [0.05*1000/51/0.5]
= 5.042
pH = 14-pOH = 14-5.042 = 8.958

10) pOH = 4.75 + log [0.05*1000/55/0.5]
= 5.01
pH = 14-pOH = 14-5.01 = 8.99

11) pOH = 4.75 + log [0.05*1000/60/0.5]
= 4.97
pH = 14-pOH = 14-4.97 = 9.03

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