Ionic EquilibriumHard
Question
Ka for HCN is 5 × 10-10 at 25oC. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 ml of 2M HCN solution is (log 2 = 0.3)
Options
A.4 ml
B.8 ml
C.2 ml
D.10 ml
Solution
Ka = 5 × 10-10 pKa = 10 log 5 = 9.3
pH = pKb + log
9 = 9.3 + log
⇒ - 0.3 = log 
0.3 = log
⇒ 
= 2 ⇒ Vml = 2 ml
pH = pKb + log
9 = 9.3 + log
⇒ - 0.3 = log 0.3 = log
⇒ = 2 ⇒ Vml = 2 mlCreate a free account to view solution
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