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

Let us first calculate `alpha` using `alpha=sqrt(K_a/c)`
`alpha=sqrt((1.8xx10^(-5))/1) =4.24xx10^(-3)`
Since , `alpha` is very much smaller than 0.1, so the assumption is valid
`therefore [H^+]=alpha=2.19xx10^(-5)`
`therefore [H^+]=sqrt(K_axxC)=sqrt(1.8xx10^(-5)xx1)=4.24xx10^(-3)`
pH=-log`(4.24xx10^(-3))=3 - log 424=3-0.6273`=2.37
Now pH on dilution = 2 × 2.37 = 4.74
`[H^+]=1.8xx10^(-5)=c alpha`
`K_a=(calpha^2)/(1-alpha)=(calpha xx alpha)/(1-alpha)=(1.8xx10^(-5)xxalpha)/(1-alpha)=1.8xx10^(-5)`
`alpha/(1-alpha)=1 therefore alpha=0.5`
`[H^+]=ac`
`c=([H^+])/alpha=(1.8xx10^(-5))/0.5=3.6xx10^(-5)`M
Let the volume to which 1 ? of 1 M acetic acid is diluted be V, then
`1 xx 1 = 3.6xx10^(-5)xxV`
`V=(1xx1)/(3.6xx10^(-5))=10^5/3.6=0.277xx10^4` litres .
You can notice that `alpha` increases on dilution and it become considerable and cannot be ignored with respect to 1.