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Question

Consider the titration of 80.0 mL of 0.147 M Ba(OH)2 by 0.588 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added.

(a) 0.0 mL

(b) 15.0 mL

(c) 29.0 mL

(d) 40.0 mL

(e) 80.0 mL

Answer 1

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

a) Mol of Ba(OH2) = 80mL x 0.147 M = 11.76 mmol

Mol of HCl = 0.588 M x 0mL = 0 mmol

Ba(OH)2 → Ba + 2OH-

[OH-] = 2 x 0.147 M = 0.294 M

pOH = -log[OH-] = 0.53

pH = 14 - pOH = 13.47

b) Ba(OH)2 + 2HCl → BaCl2 + 2H2O

11.76mmol...8.82mmol

4.41mmol...8.82mmol

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7.35 mmol

[Ba(OH)2] = 7.35 mmol / (80+15) mL = 0.077 M

[OH-] = 2 x 0.077 M = 0.154 M

pOH = 0.8125

pH = 13.2

For the next number, i didn&#039;t write my steps. The steps as same as my answer above.

c) pH = 12.77

d) pH = 7

e) 0.83