500 mL of 0.2 M aqueous solution of acetic acid is mixed with 500 mL of 0.2 M HCl at `25^(@)C`.
(i) Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution.
(ii) If 6 g of NaOH is added to the above solution, determine the final pH [Assume there is no change in volume on mixing : `K_(a) ` of acetic acid is `1.75xx10^(-5) " mol " L^(-1)`]

Explanation

(i) Millimoles of `CH_(3)CO OH = 500 xx 0.2 = 100`
Millimoles of HCl `= 500xx 0.2=100`
Final volume after mixing `= 500 + 500 = 1000 ` mL
`:. [ CH_(3)CO OH]=(100)/(1000) = 0.1 M, " " [HCl ] = (100)/(1000) = 0.1 M`
`{:(,CH_(3)CO OH ,hArr,CH_(3)CO O^(-),+,H^(+),,),("Before dissociation",0.1 M ,," "0,,0.1 M,,),(,,,,,
("from HCl"),,),("After dissociation",(0.1-x),," "x,,(0.1+x),,):}`
`K_(a) = (x(0.1+))/((0.1+x))`
As in the presence of HCl, dissociation of `CH_(3)CO OH` will be very small (due to common ion effect ), x is very very small. Hence,
`K_(a) = (x(0.1))/(0.1) = x = 1.75xx10^(-5) ` mol `L^(-1) ` (Given)
`:.` Degree of dissociation `= (x)/(0.1) = (1.75xx10^(-5))/(0.1) = 1.75xx10^(-4)= 0.00175 %`
Further, `[H^(+)]= 0.1 + x ~~ 0.1 :. pH = - log 0.1 = 1 `
(ii) 6g of NaOH = `(6)/(40) ` mole = 0.15 mole
Hence, now the equilibrium will be
`{:(,CH_(3)CO OH ,+,HCl,+,NaOH,hArr,CH_(3)CO ONa ,+,NaCl,+,H_(2)O),("Initial",0.1,,0.1,,0.15,,0,,0,,0),("At eqm.",0.05,,0,,0,,0.05,,0,,0):}`
Thus, the solution will now be 0.05 M in `CH_(3)CO OH` and 0.05 M in `CH_(3)CO ONa`, i.e., it is acidic buffer.
`pH = - log K_(a) + log. (["Salt"])/(["Acid"]) = - log (1.75xx10^(-5)) + log (0.05)/(0.05) = 4. 757`