Question

Determine the pH during the titration of 22.6 mL of 0.185 M nitric acid by 0.294 M potassium hydroxide at the following points: (1) Before the addition of any potassium hydroxide (2) After the addition of 7.10 mL of potassium hydroxide (3) At the equivalence point (4) After adding 17.0 mL of potassium hydroxide

Answer 1

1)when 0.0 mL of KOH is added

Given:

M(HNO3) = 0.185 M

V(HNO3) = 22.6 mL

M(KOH) = 0.294 M

V(KOH) = 0 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.185 M * 22.6 mL = 4.181 mmol

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

mol(KOH) = 0.294 M * 0 mL = 0 mmol

We have:

mol(HNO3) = 4.181 mmol

mol(KOH) = 0 mmol

0 mmol of both will react

remaining mol of HNO3 = 4.181 mmol

Total volume = 22.6 mL

[H+]= mol of acid remaining / volume

[H+] = 4.181 mmol/22.6 mL

= 0.185 M

use:

pH = -log [H+]

= -log (0.185)

= 0.7328

Answer: 0.733

2)when 7.1 mL of KOH is added

Given:

M(HNO3) = 0.185 M

V(HNO3) = 22.6 mL

M(KOH) = 0.294 M

V(KOH) = 7.1 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.185 M * 22.6 mL = 4.181 mmol

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

mol(KOH) = 0.294 M * 7.1 mL = 2.0874 mmol

We have:

mol(HNO3) = 4.181 mmol

mol(KOH) = 2.087 mmol

2.087 mmol of both will react

remaining mol of HNO3 = 2.094 mmol

Total volume = 29.7 mL

[H+]= mol of acid remaining / volume

[H+] = 2.094 mmol/29.7 mL

= 7.049*10^-2 M

use:

pH = -log [H+]

= -log (7.049*10^-2)

= 1.1519

Answer: 1.15

3)

This is titration of strong acid and strong base.

At equivalence point, solution would be neutral.

So, pH would be 7.00.

Answer: 7.00

4)when 17.0 mL of KOH is added

Given:

M(HNO3) = 0.185 M

V(HNO3) = 22.6 mL

M(KOH) = 0.294 M

V(KOH) = 17 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.185 M * 22.6 mL = 4.181 mmol

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

mol(KOH) = 0.294 M * 17 mL = 4.998 mmol

We have:

mol(HNO3) = 4.181 mmol

mol(KOH) = 4.998 mmol

4.181 mmol of both will react

remaining mol of KOH = 0.817 mmol

Total volume = 39.6 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.817 mmol/39.6 mL

= 2.063*10^-2 M

use:

pOH = -log [OH-]

= -log (2.063*10^-2)

= 1.6855

use:

PH = 14 - pOH

= 14 - 1.6855

= 12.3145

Answer: 12.31