What is the pH of 1 M solution of acetic acid ? To what volume one litre of this solution be diluted so that pH of the resulting solution will be twice of the original value ? `(K_(a)=1.8xx10^(-5))`

Explanation

For acetic acid `CH_3COOH + H_2O hArr CH_3COO^(-) + H_3O^+`
`[H_3O^+]=sqrt(K_a.C)=sqrt(1.8xx10^(-5)xx1)`
`=4.24xx10^(-3)`
pH=-log `(4.2xx10^(-3))` =2.372
`therefore` The pH of the diluted solution = 2 x 2.372 = 4.744
pH=4.744
-log `[H_3O^+]`=4.744
`[H_3O^+]=1.80xx10^(-5)`
Let the new concentration of `CH_3COOH` in the diluted solution be c mol `L^(-1)` .
`{:(,CH_3COOH +H_2OhArr , CH_3COO^(-)+ , H_3O^(+)),("At equ. " , c-1.8xx10^(-5),1.8xx10^(-5),1.8xx10^(-5)):}`
`K_a=([CH_3COO^-][H_3O^+])/([CH_3COOH])`
`=((1.8xx10^(-5))(1.8xx10^(-5)))/((c-1.8xx10^(-5)))=1.8xx10^(-5)`
or `(1.8xx10^(-5))/(c-1.8xx10^(-5))=1`
or `c-1.8xx10^(-5)=1.8xx10^(-5)`
or `c=1.8xx10^(-5)+1.8xx10^(-5)=3.6xx10^(-5)`
The number of moles of `CH_3COOH` before and after dilution will be the same. Let the volume of diluted solution of V litre.
`therefore` 1L of 1M `CH_3COOH` solution should be diluted to `2.78xx10^4` L.