Question

With the titration of 25.0 mL 0.216 M C2H5NH2 with 0.105 M HCl:

a. What is the pH when 10.0 mL HCl is added?

b. What is the pH at the equivalence point?

c. What is the pH when 75.0 mL HCl is added?

Answer #1

millimoles of C2H5NH2 = 25 x 0.216 = 5.4

pKb of C2H5NH2 = 3.20

a) millimoles of HCl added = 10 x 0.105 = 1.05

5.4 - 1.05 = 4.35 millimoles base left

1.05 millimoles salt formed

[base] = 4.35 / 35 = 0.124 M

[salt] = 1.05 / 35 = 0.03 M

pOH = pKb + log [salt] / [base]

pOH = 3.2 + log[0.124] / [0.03]

pOH = 3.82

pH = 14 - 3.82

pH = 10.18

b) 5.4 millimoles HCl must be added to reach equivalence point.

5.4 = V x 0.105

V = 51.43 mL

[salt] = 5.4 / 76.43 = 0.071 M

pOH = 1/2 [pKb + pKw + log C]

pOH = 1/2 [3.2 + 14 + log 0.071]

pOH = 8.02

pH = 14 - 8.02

pH = 5.98

c) millimoles of HCl added = 75 x 0.105 = 7.875

7.875 - 5.4 = 2.475 millimoles HCl left

[HCl] = 2.475 / 100 = 0.02475 M

pH = - log [H^{+}]

pH = - log [0.02475]

pH = 1.61

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pH(5.00 mL added) ------------------------------
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pH(22.0 mL added) ----------------------------
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