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Calculate the pHof a 0.01 Nsolution of acetic acid. K_(a) for CH_(3)CO OH is 1.8xx10^(-5) at 25^(@)C. |
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Answer» Solution :`CH_(3)CO OH ` ionizes as : `CH_(3)CO OH + H_(2)O hArr CH_(3)CO O^(-) + H_(3)O^(+)` Applying the law of chemical 3quilibrium, we get `K_(a) = ([CH_(3)CO O^(-)][H_(3)O^(+)])/([CH_(3)C O OH ])` But `[CH_(3)CO O^(-)]=[H_(3)O^(+)]` [`:'` one molecule of `CH_(3)COOH` gives one `CH_(3)CO O^(-)` ion and one `H^(+)` ion or `H_(3)O^(+)` ion ] `:. K_(a) = ([H_(3)O^(+)]^(2))/([CH_(3)CO OH]) or [H_(3)O^(+)]=sqrt(K_(a)[CH_(3) CO OH])` But `K_(a) = 1.8xx10^(-5) and [CH_(3)CO OH]=0.01 M = 10^(-2)M` `:. [H_(3)O^(+)]=sqrt((1.8xx10^(-5))xx(10^(-2)))=sqrt(1.8xx10^(-7))=sqrt(18xx10^(-8))=sqrt(18) xx10^(-4) g ` ions/litre `= 4.242xx10^(-4)` gions/litre `:. pH = - LOG [H_(3)O^(+)]=-log(4.242xx10^(-4))=-(0.6276-4)=3.3727=3.37`. Alternatively, as DERIVED above, `[H_(3)O^(+)]={K_(a) [CH_(3)CO OH ]}^(1//2)=(K_(a)C)^(1//2)` `:. log [H_(3)O^(+)]=1/2(log K_(a) + log C ) or - log [H_(3)O^(+)]=1/2 (- log K_(a) - log C)` i.e., `pH=1/2(pK_(a) - log C)` This formula can be used directly for finding the pH of a weak acid solution of known concentration. Similarly, for a weak base, `pOH = 1/2 (pK_(b) - log C)` |
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