dgtal wwortoncorn/24776 nd 16 HWw : 0% ix2g 12/13/17 This is a Numeric Entry que
ID: 698264 • Letter: D
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dgtal wwortoncorn/24776 nd 16 HWw : 0% ix2g 12/13/17 This is a Numeric Entry question/It is worth 3 points/You have 5 of 5 attempts remaining/There is no attempt penalty Q See page 736 14 Question (3 points) A 25.0 ml sample of a 0.110 M solution of acetic acid is titrated with a 0.142 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. The K, for acetic acid is 1.76x10-5 1st attempt Part 1 (1point) l See Periodic Table See Hint pH after 10.0 mL base added- Part 2 (1 point) pH after 200 mL base added “Explanation / Answer
millimoles of acid = 25 x 0.11 = 2.75
pKa = - log Ka = - log [1.76 x 10-5]
pKa = 4.75
1) millimoles of NaOH added = 10 x 0.142 = 1.42
2.75 - 1.42 = 1.33 millimoles acid left
1.42 millimoles salt formed
[acid] = 1.33 / 35 = 0.038 M
[salt] = 1.42 / 35 = 0.0406 M
pH = pKa + log [salt] / [acid]
pH = 4.75 + log 0.0406 / [0.038]
pH = 4.78
2) millimoles NaOH added = 20 x 0.142 = 2.84
2.84 - 2.75 = 0.09 millimoles base left
[NaOH] = 0.09 / 45 = 0.002 M
pOH = - log [OH-] = - log [0.002]
pOH = 2.70
pH = 14 - 2.70
pH = 11.3
2) millimoles NaOH added = 30 x 0.142 = 4.26
4.26 - 2.75 = 1.51 millimoles NaOH left
[NaOH] = 1.51 / 55 = 0.0274 M
pOH = - log [OH-] = - log [0.0274]
pOH = 1.56
pH = 14 - 1.56
pH = 12.44
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