Question

Consider the titration of 82.0 mL of 0.133 M Ba(OH)₂ by 0.532 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added. (with the correct sig figs)

(a) 0.0 mL
  

(b) 11.0 mL
  

(c) 29.0 mL


(d) 41.0 mL


(e) 82.0 mL

Answer 1

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

= 2*0.133 M

= 0.266 M

[H+] = 0.532 M

a)when 0.0 mL of H+ is added

Given:

M(H+) = 0.532 M

V(H+) = 0 mL

M(OH-) = 0.266 M

V(OH-) = 82 mL

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

mol(H+) = 0.532 M * 0 mL = 0 mmol

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

mol(OH-) = 0.266 M * 82 mL = 21.812 mmol

We have:

mol(H+) = 0 mmol

mol(OH-) = 21.81 mmol

0 mmol of both will react

remaining mol of OH- = 21.81 mmol

Total volume = 82.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 21.81 mmol/82.0 mL

= 0.266 M

use:

pOH = -log [OH-]

= -log (0.266)

= 0.5751

use:

PH = 14 - pOH

= 14 - 0.5751

= 13.4249

Answer: 13.42

b)when 11.0 mL of H+ is added

Given:

M(H+) = 0.532 M

V(H+) = 11 mL

M(OH-) = 0.266 M

V(OH-) = 82 mL

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

mol(H+) = 0.532 M * 11 mL = 5.852 mmol

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

mol(OH-) = 0.266 M * 82 mL = 21.812 mmol

We have:

mol(H+) = 5.852 mmol

mol(OH-) = 21.81 mmol

5.852 mmol of both will react

remaining mol of OH- = 15.96 mmol

Total volume = 93.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 15.96 mmol/93.0 mL

= 0.1716 M

use:

pOH = -log [OH-]

= -log (0.1716)

= 0.7655

use:

PH = 14 - pOH

= 14 - 0.7655

= 13.2345

Answer: 13.23

c)when 29.0 mL of H+ is added

Given:

M(H+) = 0.532 M

V(H+) = 29 mL

M(OH-) = 0.266 M

V(OH-) = 82 mL

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

mol(H+) = 0.532 M * 29 mL = 15.428 mmol

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

mol(OH-) = 0.266 M * 82 mL = 21.812 mmol

We have:

mol(H+) = 15.43 mmol

mol(OH-) = 21.81 mmol

15.43 mmol of both will react

remaining mol of OH- = 6.384 mmol

Total volume = 111.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 6.384 mmol/111.0 mL

= 5.751*10^-2 M

use:

pOH = -log [OH-]

= -log (5.751*10^-2)

= 1.2402

use:

PH = 14 - pOH

= 14 - 1.2402

= 12.7598

Answer: 12.76

d)when 41.0 mL of H+ is added

Given:

M(H+) = 0.532 M

V(H+) = 41 mL

M(OH-) = 0.266 M

V(OH-) = 82 mL

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

mol(H+) = 0.532 M * 41 mL = 21.812 mmol

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

mol(OH-) = 0.266 M * 82 mL = 21.812 mmol

We have:

mol(H+) = 21.81 mmol

mol(OH-) = 21.81 mmol

21.81 mmol of both will react to form neutral solution

hence pH of solution will be 7

Answer: 7.00

e)when 82.0 mL of H+ is added

Given:

M(H+) = 0.532 M

V(H+) = 82 mL

M(OH-) = 0.266 M

V(OH-) = 82 mL

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

mol(H+) = 0.532 M * 82 mL = 43.624 mmol

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

mol(OH-) = 0.266 M * 82 mL = 21.812 mmol

We have:

mol(H+) = 43.62 mmol

mol(OH-) = 21.81 mmol

21.81 mmol of both will react

remaining mol of H+ = 21.81 mmol

Total volume = 164.0 mL

[H+]= mol of acid remaining / volume

[H+] = 21.81 mmol/164.0 mL

= 0.133 M

use:

pH = -log [H+]

= -log (0.133)

= 0.8761

Answer: 0.876