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Nicotinic acid `(K_(a)=1.4xx10^(-5))` is represented by the formula HNiC. Calculate its percentage dissociation in a solution which contains 0.10 mole of nicotinic acid per 2.0 litre of solution. |
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Answer» Suppose degree of dissociation of nicotinic acid = `alpha` `{:(,HNiC,hArr,H^(+),+,NiC^(-)),("Initial amount",0.1 ",mole",,,,),("Amount",0.1-0.1 alpha,,0.1 alpha,,0.1 alpha),("at eqm.",=0.1 (1-alpha),,,,),("Molar conc.",=0.1(1-alpha)//2,,0.1 alpha//2,,0.1 alpha//2),("at eqm.",,,,,):}` `K_(a)=([H^(+)][NiC^(-)])/([HNiC])` or `1.4xx10^(-5)=((0.1 alpha)^(2)(0.1alpha//2))/(0.1(1-alpha)//2)` If `alpha lt lt 1`, then `1.4xx10^(-5)=((0.05 alpha)^(2))/(0.05)=0.05 alpha^(2)` or `alpha^(2)=2.8xx10^(-4)` or `alpha=1.67xx10^(-2)` % dissociation `= 1.67xx106(-2)xx100=1.67%` |
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