50mL of 0.2M aqueous solution of acetic acid is mixed with 500mL of 0.2M HCl at `25^@C`.
(i) Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution.
(ii) If 6g of NaOH is added to the above solution, determine the final pH. (Assume there is no change in volume on mixing : `K_(a) (CH_(3)COOH)=1.75 xx 10^(-5)"mol/L")`

Explanation

(i) As equal volumes of `CH_(3)COOH and HCl` are mixed, molarity of each will be 0.1mole/L or 0.1M.
`underset(0.1(1-x))overset(0.1 M)((CH_(3)COOH)) Leftrightarrow underset(0.1x)(CH_(3)COO^(-))+underset((0.1x+0.1)) (H^+)`
where x is the degree of dissociatioin of `CH_(3)COOH` in the presence of HCl.
`K_(a)=(0.1 x xx 0.1)/(0.1 (1-x))=0.1x ("x is a very small")`
`x=(1.75 xx 10^(-5))/(0.1) =1.75 xx 10^(-4)`
`pH =-log [H^+]=-log (0.1)=1, (H^+ "ions from "CH_(3)COOH "neglected")`
(ii) Molarity of NaOH `=6/(40)" mole/L=0.15M"`
0.15 mole of NaOH when added to this acid mixture, will neutralise 0.1 mole of HCl and 0.05 mole of `CH_(3)COOH`. The resulting solution will be 0.05 M in `CH_(3)COOH` and 0.05M in `CH_(3)COONa` which is a buffer solution.
`pH +pK_(a)+log (["salt"])/(["acid"])`
`pH =-log (1.75 xx 10^(-5))+log ""(0.05)/(0.05)`
`pH =4.757`