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

Case I:
`{:(,CH_(3)COOHhArr,CH_(3)COO^(-)+,H^(+)),("Conc. Before dissociation",C,0,0),("Conc. After dissociation,1-alpha,alpha,alpha):}`
`[H^(+)]=Calpha=Csqrt(((K_(a))/(C )))sqrt((K_(a)C))`
`=sqrt((1.8xx10^(-5)xx1))`
`=sqrt((18xx10^(-6)))=4.24xx10^(-3)M`
`:. pH= -log H^(+)= -log (4.24xx10^(-3))`
`pH= 2.3724`
Case II: New pH is given as `= 2.3724xx2=4.7448` Let new conc. be `C_(1)` and degree of dissociation be `alpha_(1)`
`:. -log[H^(+)]=4.7448`
`:. [H^(+)]=1.8xx10^(-5)`
or `C_(1)alpha_(1)=1.8xx10^(-5)`
Now again `K_(a)= ([H^(+)][CH_(3)COO^(-)])/([CH_(3)COOH])`
`K_(a)=(C_(1)alpha_(1)xxC_(1)alpha_(1))/(C_(1)(1-alpha_(1)))=(C_(1)alpha_(1).alpha_(1))/((1-alpha_(1)))`
`:. 1.8xx10^(-5)=(1.8xx10^(-5)xxalpha_(1))/((1-alpha_(1)))`
`alpha_(1)=0.5`
Now `C_(1)alpha_(1)=1.8xx10^(-5)`
`C_(1)= (1.8xx10^(-5))/(alpha_(1))=(1.8xx10^(-5))/(0.5)`
`=3.6xx10^(-5)M`
Let 1 litre of conc. solution be diluted V litre. Eq. of dilute solution = Eq. of concentrated solution
`3.6xx10^(-5)xxV=1xx1`
`:. V=(1)/(3.6xx10^(-5))=2.77xx10^(4)`litre
Note: In II case `alpha_(1)` comes `0.5` by `K_(a)=(C_(1)alpha_(1)^(2))/((1-alpha_(1)))` and thus, it is not advisable to assume `(1-alpha_(1))~~1`.