`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))`

Answer

D

Explanation

`20 mL` of `0.2M` sodium hydroxide is added to `50 mL` of `0.2M` acetic acid to acid to given `70 mL` of the solution. An acidic buffer is formed.
`{:(NaOH+,CH_(3)COOHrarr,CH_(3)COONa+,H_(2)O,),(("millimoles"),,,,),(20xx0.2,50xx0.2,,,),(=4,=10,0,0,),(0,10-4=6,4,4,):}`
`["Acid"] = (7)/(70) ["Salt"] = (4)/(70)`
`pH = pK_(a) +"log" (["Salt"])/(["Acid"])`
`=- log (1.8 xx 10^(-5)) +log"(4)/(70) xx (70)/(6)`
`= 4.74 + "log"(2)/(3) = 4.56`
Rule ABA: When `NaOH` is added, the amount of acid decreases because hydroxide ions react with hydrogen ions to form undissociated water. Therefore, the amount of acetate increases.
Initial millimoles of acid `= 6`
New millimoles of acid `= 6 - (0.2V)`
New `["Acid"] = (6-(0.2V))/(70)`
Initial millimoles of salt `= 4`
New millimoles of salt `= 4 + (0.2 V)`
New `["Salt"] = (4+(0.2V))/(70)`
`pH = pK_(a) "log"(["Salt"])/(["Acid"])`
or `4.74 =- log (1.8 xx 10^(-5))`
`+log {(4+(0.2V))/(70)xx(70)/(6-(0.2V))}`
`- log (1.8 xx 10^(-5)) ~~ 4.74`
`:. 4.74 = 4.74 + "log" (4+(0.2V))/(6-(0.2V))`
or `0 = "log" (4+(0.2V))/(6-(0.2V))`
or `1 = (4+(0.2V))/(6-(0.2V))`
or `4 +(0.2V) = 6 - (0.2V)`
or `0.4 V = 2`
or `V =5mL`