What is the `pH` of `1M` 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) = 1.8 xx 10^(-5)`

Explanation

`{:(" "H_(3)C COOH" "H_(2)O" "hArr" " H_(3)C COO^(-)" + "H_(3)O^(+)),("Initial 1M 0 0"),(" "-xM" "xM" "xM),("Final "bar(" "(1-x)M" "x" "x" ")):}`
`K_(a) = (x^(2))/(1-x)approx (x^(2))/1 :. x = sqrt(K_(a)) = 4.2 xx10^(-3) = [ H_(3)O^(+)] `
`pH = - log [H_(3)O^(+)] = - log { 4.2 xx10^(-3)} = 3 - 4.2 = 2.37 `
Now , let 1 L M AcOH solution be diluted to V to double the pH and the conc. of diluted solution be C.
New `pH = 2 xx Old pH = 2 xx 2.37 = 4.74 `
` pH = - log [H_(3)O^(+)] = 4.74`
`pH = -log [H_(3)O^(+)] = 4.74 `
` :. [H_(3)O^(+)] = 1.8 xx10^(-5)`
`{:(" "H_(3)C COOH+H_(2)O hArr H_(3)C COO^(-)" "+ " "H_(3)O^(+)),("Initial C 0 0"),(" "ul(" "-1.8xx10^(-5) " " 1.8xx10^(-5)" "1.8xx10^(-5))),("Final " C-1.8xx10^(-5)" "1.8xx10^(-5)" "1.8xx10^(-5)):}`
`K_(a)= ([CH_(3)COO^(-)]xx[H_(3)O^(+)])/([CH_(3)COOH])`
` 1.8 xx 10^(-5) = (1.8 xx10^(-5)xx1.8 xx10^(-5))/(C-1.8xx10^(-5))`
` :. C = 3.6 xx10^(-5) L `
On dilution ,
`M_(1)V_(1) = M_(2)V_(2)`
` 1 M xx 1 L = 3.6 xx10^(-5) L xx V_(2)`
` :. V_(2) = 2.78 xx 10^(4) L `