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Calculate the amount of acetic acid pres...

Calculate the amount of acetic acid present in one litre, having degree of dissociation equal 1% and `K_(a) = 1.8 xx 10^(-5)`.

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`{:(,CH_(3)COOH,hArr,CH_(3)OO^(-),+,H^(+)),("Initially",1,,0,,0),("After dissociation",1-alpha,,alpha,,alpha):}`
wehre `.alpha.` is the degree of dissocation, if C is the concentration of acetic acid C mol `"litre"^(-1)`
`k_(a) = ([ CH_(3)COO^(-)][H^(+)])/(CH_(3)COOH)`
`= ( alpha C. alpha C )/( C ( 1-alpha))=( alpha^(2) C^(2))/( c(1-alpha))`
`rArr = ( alpha^(2) C)/( ( 1- alpha))`
If `.alpha.` is very small, then `1- alpha =1`
Therefore, `k_(a) = alpha^(2) C`
Then, `1.8 xx 10^(-5) = Cxx [ ( 1)/( 100) ] ^(2)`
`C = 0.18 `mol. `"litre"^(-1)`
Molecular mass of `CH_(3) COOH = 60g`
Therefore, 1 litre solution contains 0.18 moles
`= 0.18 xx 60 = 10.8 g`
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