Ionic EquilibriumHard
Question
An ammonia - ammonium chloride buffer has a pH value of 9 with [NH3] = 0.25. What will be the newpH if 500 ml 0.1 M KOH is added to 200 ml buffer solution (Kb = 2 × 10-5) [log 2 = 0.3]
Options
A.8.4
B.9.6
C.5.6
D.4.4
Solution
5 = pKb + log
[NH4+] = 0.5
[NH3] = 0.25
m moles of KOH added = 50
NH4+ + OH- → NH3 + H2O
100 50 50
50 0 100
pOH = pKb + log
pOH = 4.7 - 0.3 ⇒ 4.4
pH = 9.6
[NH4+] = 0.5
[NH3] = 0.25
m moles of KOH added = 50
NH4+ + OH- → NH3 + H2O
100 50 50
50 0 100
pOH = pKb + log
pOH = 4.7 - 0.3 ⇒ 4.4
pH = 9.6
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