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K(a) for HA is 4.9 xx 10^(-8). After mak...

`K_(a)` for `HA` is `4.9 xx 10^(-8)`. After making the necessary approximation, calculate for its decimolar solution,
a. `%` dissociation b. `overset(Θ)OH` concentration
c. `pH`

Text Solution

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`{:(HA,hArr,H^(+),+,A^(-)),(C,,0,,0),(C(1-alpha),,alpha,,alpha):}`
`K_(a) = ([H^(+)][A^(-)])/([HA]) rArr K_(a) = (Calpha .Calpha)/(C(1-alpha)) = (Calpha^(2))/((1-alpha))approx Calpha^(2)`
(i) ` :. alpha = sqrt((K_(a))/C) = sqrt((4.9 xx10^(-8))/(1//10)) (C=1//10 M)`
` = 7 xx10^(-4) = 0.07 % `
(ii) `[H^(+)] = Calpha = 1/10 xx 7 xx10^(-4) = 7 xx10^(-5) ` mol/L
(iii) `[H^(+)] [OH^(-)] = 1xx10^(-14)`
` :. [OH^(-)] = (10^(-14))/(7xx10^(-5)) = 1.43 xx10^(-10) ` mol/L
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