Consider a weak monobasic acid HA having a concentration C with degree of dissociation, α<<1.. The pH of this weak acid in terms of acid dissociation constant Ka and concentration C is :
A
−log10(√KaC)
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B
−log10√Ka.C
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C
pH=−log10√Ka.C2
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D
pH=−log10√K2a.C
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Solution
The correct option is B−log10√Ka.C For weak acid HA with concentration ‘C′, ionization constant Ka and degree of dissociation (α<<1) In aqueous solution, HA(aq)⇌H+(aq)+A−(aq) At t=0C00 EquilibriumC−CαCαCα Ka=Cα21−α As α<<1,(1−α)≈1
Ka=Cα2
As H+ concentration is Cα By multiplying C on both sides, we get: Ka×C=Cα2×C=Ka×C=(Cα)2Ka×C=[H+]2 [H+]=√Ka.C pH=−log10√Ka.C