Calculate the pH of each of the following solution
(i) 100 ml of 0.1 M `CH_(3)COOH` mixed with 100 ml of 0.1 M NaOH.
(ii) 100 ml of 0.1 M `CH_(3)COOH` mixed with 50 ml of 0.1 m NaOH
(iii) `50 ml of 0.1 M CH_(3)COOH` mixed with 100 ml of 0.1 M NaOH.
`K_(a)(CH_(3)COOH)=1.8xx10^(-5)`

Explanation

(a) `underset("1 mole")(CH_(3)COOH)+underset("1 mole")(NaOH)toCH_(3)underset("1 mole")(COONa)+underset("1 mole")(H_(2)O)`
m. moles of `Ch_(3)COOH=0.1xx100=10`
m moles of `NaOH=100xx0.1=10`
Therefore, m. moles of `CH_(3)COONa` is 10
Now in the solution only salt `CH_(3)COONa` is present, therefore, pH of solution can be calculated by the hydrolysis of `CH_(3)COONa`
`CH_(3)COONa` is a salt of weak acid with strong base
`pH=7+1/2pK_(a)+1/2logC`
`=7+1/2xx4.75+1/2log((10)/(200))" "(pK_(a)=-log1.8xx10^(-5))`
`=8.72" "pK_(a)=4.75`
(b) `CH_(3)COOH+NaOHtoCH_(3)COONa+H_(2)O`
`{:("Initial m moles",10,5,0,0),("m moles after reaction",10-5,0,5,5):}`
Since in the solution, mixture of `CH_(3)COOH and CH_(3)COONa` is present therefore, solution is an acidic buffer and pH can be calculated as
`pH=pK_(a)+log""(["salt"])/(["acid"])pH=4.75+log""(((5)/(200))/((5)/(200)))pH=4.75`
(c ) `CH_(3)COOH+NaOHto CH_(3)COONa+H_(2)O`
`{:("Initial m moles",5,10,0,0),("m moles after reaction",0,(10-5),5,5):}`
Now in the solution strong base NaOH and salt of weak aid with strong base is present We can assume that `OH^(-)` ions concentration from the hydrolysis of salt is negligible because hydrlysis of `CH_(3)COONa` is suppressed by the presence of strong base NaOH.
`underset(("from salt"))(CH_(3)COO^(-))+H_(2)OhArrCH_(3)COOH+OH^(-)`
`therefore[OH^(-)]=([5])/([150])=((1)/(30))`
`pOH=log[OH^(-)]=-log((1)/(30))=log30=1.4771`
`pH+pOH=14`
`pH=14-1.4771=12.52`