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For the titration of 75 mL of 0.10 M acetic acid with 0.10 M NaOH, calculate...

For the titration of 75 mL of 0.10 M acetic acid with 0.10 M NaOH, calculate the pH. For acetic acid, HC2H3O2, Ka = 1.8 x 10-5. (a) before the addition of any NaOH solution. (b) after 25 mL of the base has been added. (c) after half of the HC2H3O2 has been neutralized. (d) at the equivalence point.

Solutions

Expert Solution

millimoles of acid = 75 x 0.1 = 7.5

pKa of acid = 4.74

a) before the addition of any NaOH solution

pH = 1/2 [pKa - logC]

pH = 1/2 [4.74 -log 0.1]

pH = 2.87

b) 25.00 mL of 0.100 M NaOH

millimoles of NaoH = 25 x 0.1 = 2.5

HA   +   NaOH ------------------> NaA + H2O

7.5           2.5                            0            0

5.0          0                               2.5         2.5

pH = pKa + log [NaA/ HA]

pH = 4.74 + log (2.5 / 5.0)

pH = 4.44

(c) after half of the HC2H3O2 has been neutralized

pH = pKa

pH = 4.74

(d) at the equivalence point.

volume of base needed = 75 mL

concentration of salt = 7.5 / (75 + 75) = 0.05 M

pH = 7 + 1/2 [pKa + logC]

pH = 7 + 1/2 [4.74 + log 0.05]

pH = 8.72


queen_honey_blossom answered 2 years ago
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