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The Ka of propionic acid is 1.34 xx 10^(-5). What is the pH of a solution containing 0.5 M propionic and 0.5 M sodium proportionate ? What happens to the pH of the solution when volume is doubled by adding water ? |
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Answer» Solution :`Ka" of propionic acid "=1.34xx10^(-5)` `therefore""pK_(a)=-logK_(a)=-log(1.34xx10^(-5))` `=4.87` By Herderson - Hasselbalch EQUATION `pH=pK_(a)+log.(["salt"])/(["acid"])` `=4.87+log""(0.5)/(0.5)` `pH=4.87` 87 Alternative solution : The dissociation equilibrium of propionic acid will be 2)H_(5)COOH rarr C_(2)H_(5)COO^(-)+H^(+)` `K_(a)=([C_(2)H_(5)COO^(-)][H^(+)])/([C_(2)H_(5)COOH])=(0.5xx[H^(+)])/(0.5)` `=[H^(+)]` `therefore""pH=-log[H^(-)]` `=-logK_(a)=-log(1.34xx10^(-5))` `therefore pH=4.87` |
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