

InterviewSolution
Saved Bookmarks
1. |
`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» Correct Answer - D `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` |
|