`20 mL` of `0.2M` sodium hydroxide is added to `50 mL` of `0.2 M `acetic acid to give `70 mL` of the solution. What is the `pH` of this solution. Calculate the additional volume of `0.2M NaOh` required to make the `pH` of the solution `4.74`. (Ionisation constant of `CH_(3)COOh` is `1.8 xx 10^(-5))`

Explanation

We know, number of millimoles `="molarity "xx " volume (mL) "`
millimoles of NaOH `=0.2 xx 20=4`
Millimoles of `CH_(3)COOH=0.2 xx 50=10`
Millimoles of `CH_3COOHNa` produced =4
millimoles of `CH_3COOH` remained =10-4=6
Now, `pH=pK_a+log ""("millimoles of salt")/("millimoles of acid")`
`=-log (1.8 xx 10^(-5)) +log ""4/6`
=4.56
Let the additional volume of NaOH be v mL to make the pH 4.74
Millimoles of added NaOH=0.2v
Since 0.2v m.m. of NaOH will neutralise 0.2v m.m. of `CH_3COOH and" form 0.2v m.m. of "CH_(3)COONa`,
millimoles of `CH_3COO=(6-0.2v)`
millimoles of `CH_3COONa=(4+0.2v)`
Again, using Equation 4 (b), we get
`pH =pK_a+log ""(4+0.2v)/(6-0.2v)`
`4.74=-log (1.86 xx 10^(-5))+log ""(4+0.2v)/(6-0.2v)`
`4.74=4.7447+log ""(4+0.2v)/(6-0.2v)`
`-0.0047=log""(4+0.2v)/(6-0.2v)`
Taking antilog `(6-0.2v)/(4+0.2v)=1.011` v=4.86mL