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Consider the titration of a 24.0ml sample of 0.105 M CH3COOH with .130 M Naoh. The...

Consider the titration of a 24.0ml sample of 0.105 M CH3COOH with .130 M Naoh. The La of acetic acid is 1.70x10^-5. what is the a)initial ph? b) the ph at 6.00 of added base? c) the ph at the equivalence point?

Solutions

Expert Solution

Ka = 1.70 x 10-5

pKa = - log Ka = - log [1.70 x 10-5]

pKa = 4.77

millimoles of CH3COOH = 24 x 0.105 = 2.52

a) initially only CH3COOH present so

pH = 1/2 [pKa - logC]

pH = 1/2 [4.77 - log 0.105]

pH = 2.87

b) millimoles of NaOH added = 6.0 x 0.130 = 0.78

2.52 - 0.78 = 1.74 millimoles acid left

0.78 millimoles salt formed

total volume = 24 + 6 = 30 mL

[acid] = 1.74 / 30 = 0.058M

[salt] = 0.78/30 = 0.026M

mixture will act as acidic buffer

pH = pKa + log [salt] / [acid]

pH = 4.77 + log [0.026] / [0.058]

pH = 4.72

c) at equivalence all acid must be convert to salt.

2.52 = V x 0.13

V = 19.38 mL NaOH required

total volume = 24 + 19.38 = 43.38 mL

[salt] = 2.52 / 43.38 = 0.0581 M

pH = 1/2 [pKw + pKa + log C]

pH = 1/2 [14 + 4.77 + log 0.0581]

pH = 8.77

queen_honey_blossom answered 2 years ago
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