Consider a weak monoacidic base BOH having a concentration C with dissociation α<<1. The pH of this weak base in terms of base dissociation constant Kb and concentration C is :
A
14+log10(√KbC)
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B
14+log10√Kb.C
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C
14−log10√Kb.C
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D
14−log10(√KbC)
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Solution
The correct option is B14+log10√Kb.C For weak base BOH with concentration ‘C′, ionization constant Kb and α<<1. In aqueous solution, BOH(aq)⇌B+(aq)+OH−(aq) At t=0C00 At t=t(eq)C−CαCαCα
Kb=Cα21−α
As α<<1(1−α)≈1
Kb=Cα2 Multiplying both sides by C Kb×C=Cα2×CKb×C=[OH−]2 [OH−]=√Kb.C pOH=−log10√Kb.CpH+pOH=14pH=14+log10√Kb.C