Question

# Find the dissociation constant Ka of a weak monobasic acid which is 3.5% dissociated in a M10 solution at 25∘C

A
0.5×103
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B
2.62×106
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C
4.2×106
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D
1.27×104
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Solution

## The correct option is B 1.27×10−4 HA(aq.)⇌ H+(aq.)+A−(aq.)Initial: C 0 0Equilibrium:(C−Cα) Cα CαC=110=0.1Mα=3.5100=0.035Ka=[H+][A−][HA]=Cα21−αKa=0.1×(0.035)21−0.035Ka=1.269×10−4 M Theory: Relation between Ka and Kb: Consider this weak acid reaction of HA HA(aq)+H2O(l)⇌ H3O+(aq)+A−(aq) Ka=[H3O+][A][HA] Consider this weak base reaction of the conjugate base of HA. I.e., A− A−(aq)+H2O(l)⇌ HA(aq)+OH−(aq) Kb=[OH−][HA][A] Ka×Kb=[H3O+][A][HA]×[OH−][HA][A] Ka×Kb=[H3O+][OH−]=Kw or Ka×Kb=[H+][OH−]=Kw=10−14 For any acid-base conjugate pair, Ka×Kb=Kw

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