`500mL` of `0.2M` aqueous solution of acetic acid is mixed with `500mL` of `0.2HCI` at `25^(@)C`.
a. Calculate the degree of dissociation of acetic acid in the resulting solution and `pH` of the folution.
b. If `6g` of `NaOH` is added to the above solution determine the final `pH.[K_(a)` of `CH_(3)COOH =2 xx 10^(-5)`.

Explanation

(a) Meq. Of `CH_(3)COOH=500xx0.2=100`
Meq. Of `HCI=500xx0.2=100`
`:. [HCI]=(100)/(1000)=0.1, [CH_(3)COOH]=(100)/(1000)=0.1`
For `CH_(3)COOH:` `{:(,CH_(3)COOHhArr,CH_(3)COO^(-)+,H^(+)),("Before dissociation",0.1,0,0.1("from HCI")),("After dissociation", (0.1x),x,(0.1+x)):}`
Due to common ion effect, dissociation of `CH_(3)COOH` is very small in presence of HCI.
Therefore `(0.1+x)=0.1 and (0.1-x)=0.1`.
`:. K_(a)=(x xx0.1)/(0.1)`
`:. x=K_(a)=1.75xx10^(-5)`
Thus degree of dissociation `alpha= (x)/(0.1)=(1.757xx10^(-5))/(0.1)`
`=1.75xx10^(-4)`
`=0.000175=0.0175%`
Also `[H^(+)]=0.1+x=0.1` , `(.: xltltalpha1)`
`:. pH= -log[H^(+)]= -log[0.1]=1`
(b) Meq. of NaOH mole of NaOH added `= (6)/(40)=0.15`
Therefore new equilibrium will have
`{:(CH_(3)COOH+,HCI+,NaOHrarr,CH_(3)COONa+,NaCI+,H_(2)O),(0.1,0.1,0.15,0,0,0),(0.5,0,0,0.05,0,0):}`
Thus the solution will act as buffer having `[CH_(3)COOH]`
`=(0.5)/(1000)` and `[CH_(3)COONa]=(0.5)/(1000)`
Thus, `pH= -log K_(a)+log ((["Salt"])/(["Acid"]))`
`= -log 1.75xx10^(-5)+log (([0.05//1000])/([0.05//1000]))`
`pH=4.757`