Question

Determine the pH during the titration of 17.3 mL of 0.296 M nitric acid by 0.242 M barium hydroxide at the following points: (1) Before the addition of any barium hydroxide (2) After the addition of 5.30 mL of barium hydroxide (3) At the equivalence point (4) After adding 12.7 mL of barium hydroxide

Answer 1

1) Before addition of barium hydroxide

HNO3 = [H+] = 0.296 M

pH = -log[H+]

   = -log0.296

   = 0.529

2) After the addition of 5.30 mL of barium hydroxide

2 HNO3(aq) + Ba(OH)2(aq) ----> Ba(NO3)2(aq) + 2H2O(l)

2 mol HNO3(aq) = 1 mol Ba(OH)2(aq)

no of mol of HNO3 taken = M*V

                       = 0.296*17.3

                       = 5.121 mmol

no of mol of Ba(OH)2 taken = M*V

                           = 5.3*0.242

                           = 1.283 mmol

no of mol of excess HNO3 = 5.121-(2*1.283) = 2.555 mmol

concentration of excess HNO3 = 2.555/(17.3+5.3)

                             = 0.113 M

pH = -log[H+]

   = -log0.113

   = 0.95

3) At the equivalence point

2 HNO3(aq) + Ba(OH)2(aq) ----> Ba(NO3)2(aq) + 2H2O(l)

Both HNO3, Ba(OH)2 are strong acid,bases.so that, at the equivalence point pH = 7

4)

After the addition of 5.30 mL of barium hydroxide

2 HNO3(aq) + Ba(OH)2(aq) ----> Ba(NO3)2(aq) + 2H2O(l)

2 mol HNO3(aq) = 1 mol Ba(OH)2(aq)

no of mol of HNO3 taken = M*V

                       = 0.296*17.3

                       = 5.121 mmol

no of mol of Ba(OH)2 taken = M*V

                           = 12.7*0.242

                           = 3.0734 mmol

no of mol of excess Ba(OH)2 = (2*3.0734)-5.121 = 1.026 mmol

concentration of excess [OH-] = (2*1.026)/(17.3+12.7)

                             = 0.0684 M

pOH = -log[OH-]

   = -log0.0684

   = 1.165

pH = 14-1.165 = 12.835