Ionic EquilibriumEasy
Question
pH of a solution made by mixing 50 mL of 0.2 M NH4Cl and 75 mL of 0.1 M NaOH is :-
[pKb of NH3(aq) = 4.74. log 3 = 0.47]
Options
A.7.02
B.13.0
C.7.02
D.9.73
Solution
NH4Cl + NaOH → NH4OH + NaCl
mmol 50 × 0.2 75 × 0.1
= 10 = 7.5
mmol 10 - 7.5 - 7.5
= 2.5
⇒ This will result in a basic buffer.
⇒ pOH = pKb + log (Salt/Base)
= 4.74 + log
= 4.27
⇒ pH = 14 - 4.27 = 9.73
mmol 50 × 0.2 75 × 0.1
= 10 = 7.5
mmol 10 - 7.5 - 7.5
= 2.5
⇒ This will result in a basic buffer.
⇒ pOH = pKb + log (Salt/Base)
= 4.74 + log
= 4.27⇒ pH = 14 - 4.27 = 9.73
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