Question

Determine the pH during the titration of 22.7 mL of 0.236 M nitric acid by 7.88×10-2 M barium hydroxide at the following points: (1) Before the addition of any barium hydroxide (2) After the addition of 17.0 mL of barium hydroxide (3) At the equivalence point (4) After adding 42.2 mL of barium hydroxide

Answer 1

1 Ba(OH)2 has 2 OH-

So,

[OH-] = 2*[Ba(OH)2]

= 2*7.88*10^-2 M

= 0.1576 M

[H+] = [HNO3] = 0.236 M

1)

1)when 0.0 mL of OH- is added

Given:

M(H+) = 0.236 M

V(H+) = 22.7 mL

M(OH-) = 0.1576 M

V(OH-) = 0 mL

mol(H+) = M(H+) * V(H+)

mol(H+) = 0.236 M * 22.7 mL = 5.3572 mmol

mol(OH-) = M(OH-) * V(OH-)

mol(OH-) = 0.1576 M * 0 mL = 0 mmol

We have:

mol(H+) = 5.357 mmol

mol(OH-) = 0 mmol

0 mmol of both will react

remaining mol of H+ = 5.357 mmol

Total volume = 22.7 mL

[H+]= mol of acid remaining / volume

[H+] = 5.357 mmol/22.7 mL

= 0.236 M

use:

pH = -log [H+]

= -log (0.236)

= 0.6271

Answer: 0.627

2)when 17.0 mL of OH- is added

Given:

M(H+) = 0.236 M

V(H+) = 22.7 mL

M(OH-) = 0.1576 M

V(OH-) = 17 mL

mol(H+) = M(H+) * V(H+)

mol(H+) = 0.236 M * 22.7 mL = 5.3572 mmol

mol(OH-) = M(OH-) * V(OH-)

mol(OH-) = 0.1576 M * 17 mL = 2.6792 mmol

We have:

mol(H+) = 5.357 mmol

mol(OH-) = 2.679 mmol

2.679 mmol of both will react

remaining mol of H+ = 2.678 mmol

Total volume = 39.7 mL

[H+]= mol of acid remaining / volume

[H+] = 2.678 mmol/39.7 mL

= 6.746*10^-2 M

use:

pH = -log [H+]

= -log (6.746*10^-2)

= 1.171

Answer: 1.17

3)

This is titration of strong acid and strong base.

At equivalence solution would be neutral and hence pH would be 7.00

Answer: 7.00

4)when 42.2 mL of OH- is added

Given:

M(H+) = 0.236 M

V(H+) = 22.7 mL

M(OH-) = 0.1576 M

V(OH-) = 42.2 mL

mol(H+) = M(H+) * V(H+)

mol(H+) = 0.236 M * 22.7 mL = 5.3572 mmol

mol(OH-) = M(OH-) * V(OH-)

mol(OH-) = 0.1576 M * 42.2 mL = 6.6507 mmol

We have:

mol(H+) = 5.357 mmol

mol(OH-) = 6.651 mmol

5.357 mmol of both will react

remaining mol of OH- = 1.294 mmol

Total volume = 64.9 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.294 mmol/64.9 mL

= 1.993*10^-2 M

use:

pOH = -log [OH-]

= -log (1.993*10^-2)

= 1.7005

use:

PH = 14 - pOH

= 14 - 1.7005

= 12.2995

Answer: 12.30