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A `0.02 M` solution of pyridinium hydrochloride has `pH=3.44`. Calculate the ionization constant of pyridine. |
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Answer» Correct Answer - `K_(b) = 1.5 xx 10^(-9)` pH = 3.44 We know that, `pH=-log[H^(+)]` `therefore[H^(+)]=3.63xx10^(-4)` Then, `K_(h)=((3.63xx10^(-4))^(2))/(0.02)" "(because"concentration = 0.02M")` `rArrK_(h)=6.6xx10^(-6)` Now, `K_(h)=(K_(w))/(K_(a))` `rArrK_(a)=(K_(w))/(K_(h))=(10^(-14))/(6.6xx10^(-6))` `=1.51xx10^(-9)` |
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