Question

A 50.0 mL solution of 0.118 M KOH is titrated with 0.236 M HCl. Calculate the pH of the solution after the addition of each of the given amounts of HCl.

0.00 mLpH=

5.00 mLpH=

12.5 mLpH=

18.0 mLpH=

24.0 mLpH=

25.0 mLpH=

26.0 mLpH=

30.0 mLpH=

Answer 1

1)when 0.0 mL of HCl is added

Given:

M(HCl) = 0.236 M

V(HCl) = 0 mL

M(KOH) = 0.118 M

V(KOH) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.236 M * 0 mL = 0 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.118 M * 50 mL = 5.9 mmol

We have:

mol(HCl) = 0 mmol

mol(KOH) = 5.9 mmol

0 mmol of both will react

remaining mol of KOH = 5.9 mmol

Total volume = 50.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 5.9 mmol/50.0 mL

= 0.118 M

use:

pOH = -log [OH-]

= -log (0.118)

= 0.9281

use:

PH = 14 - pOH

= 14 - 0.9281

= 13.0719

2)when 5.0 mL of HCl is added

Given:

M(HCl) = 0.236 M

V(HCl) = 5 mL

M(KOH) = 0.118 M

V(KOH) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.236 M * 5 mL = 1.18 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.118 M * 50 mL = 5.9 mmol

We have:

mol(HCl) = 1.18 mmol

mol(KOH) = 5.9 mmol

1.18 mmol of both will react

remaining mol of KOH = 4.72 mmol

Total volume = 55.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 4.72 mmol/55.0 mL

= 8.582*10^-2 M

use:

pOH = -log [OH-]

= -log (8.582*10^-2)

= 1.0664

use:

PH = 14 - pOH

= 14 - 1.0664

= 12.9336

3)when 12.5 mL of HCl is added

Given:

M(HCl) = 0.236 M

V(HCl) = 12.5 mL

M(KOH) = 0.118 M

V(KOH) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.236 M * 12.5 mL = 2.95 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.118 M * 50 mL = 5.9 mmol

We have:

mol(HCl) = 2.95 mmol

mol(KOH) = 5.9 mmol

2.95 mmol of both will react

remaining mol of KOH = 2.95 mmol

Total volume = 62.5 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.95 mmol/62.5 mL

= 4.72*10^-2 M

use:

pOH = -log [OH-]

= -log (4.72*10^-2)

= 1.3261

use:

PH = 14 - pOH

= 14 - 1.3261

= 12.6739

4)when 18.0 mL of HCl is added

Given:

M(HCl) = 0.236 M

V(HCl) = 18 mL

M(KOH) = 0.118 M

V(KOH) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.236 M * 18 mL = 4.248 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.118 M * 50 mL = 5.9 mmol

We have:

mol(HCl) = 4.248 mmol

mol(KOH) = 5.9 mmol

4.248 mmol of both will react

remaining mol of KOH = 1.652 mmol

Total volume = 68.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.652 mmol/68.0 mL

= 2.429*10^-2 M

use:

pOH = -log [OH-]

= -log (2.429*10^-2)

= 1.6145

use:

PH = 14 - pOH

= 14 - 1.6145

= 12.3855

5)when 24.0 mL of HCl is added

Given:

M(HCl) = 0.236 M

V(HCl) = 24 mL

M(KOH) = 0.118 M

V(KOH) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.236 M * 24 mL = 5.664 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.118 M * 50 mL = 5.9 mmol

We have:

mol(HCl) = 5.664 mmol

mol(KOH) = 5.9 mmol

5.664 mmol of both will react

remaining mol of KOH = 0.236 mmol

Total volume = 74.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.236 mmol/74.0 mL

= 3.189*10^-3 M

use:

pOH = -log [OH-]

= -log (3.189*10^-3)

= 2.4963

use:

PH = 14 - pOH

= 14 - 2.4963

= 11.5037

6)when 25.0 mL of HCl is added

Given:

M(HCl) = 0.236 M

V(HCl) = 25 mL

M(KOH) = 0.118 M

V(KOH) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.236 M * 25 mL = 5.9 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.118 M * 50 mL = 5.9 mmol

We have:

mol(HCl) = 5.9 mmol

mol(KOH) = 5.9 mmol

5.9 mmol of both will react to form neutral solution

hence pH of solution will be 7

7)when 26.0 mL of HCl is added

Given:

M(HCl) = 0.236 M

V(HCl) = 26 mL

M(KOH) = 0.118 M

V(KOH) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.236 M * 26 mL = 6.136 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.118 M * 50 mL = 5.9 mmol

We have:

mol(HCl) = 6.136 mmol

mol(KOH) = 5.9 mmol

5.9 mmol of both will react

remaining mol of HCl = 0.236 mmol

Total volume = 76.0 mL

[H+]= mol of acid remaining / volume

[H+] = 0.236 mmol/76.0 mL

= 3.105*10^-3 M

use:

pH = -log [H+]

= -log (3.105*10^-3)

= 2.5079

8)when 30.0 mL of HCl is added

Given:

M(HCl) = 0.236 M

V(HCl) = 30 mL

M(KOH) = 0.118 M

V(KOH) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.236 M * 30 mL = 7.08 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.118 M * 50 mL = 5.9 mmol

We have:

mol(HCl) = 7.08 mmol

mol(KOH) = 5.9 mmol

5.9 mmol of both will react

remaining mol of HCl = 1.18 mmol

Total volume = 80.0 mL

[H+]= mol of acid remaining / volume

[H+] = 1.18 mmol/80.0 mL

= 1.475*10^-2 M

use:

pH = -log [H+]

= -log (1.475*10^-2)

= 1.8312