Question

17.3 Mastery #1 Q3
pH Titration pH's Strong Acid + Strong Bases

Determine the pH during the titration of 17.5 mL of 0.341 M hydroiodic acid by 0.326 M sodium hydroxide at the following points:

(1) Before the addition of any sodium hydroxide ___

(2) After the addition of 9.15 mL of sodium hydroxide ___

(3) At the equivalence point ___

(4) After adding 22.5 mL of sodium hydroxide ___

Answer 1

1)when 0.0 mL of NaOH is added

Given:

M(HI) = 0.341 M

V(HI) = 17.5 mL

M(NaOH) = 0.326 M

V(NaOH) = 0 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.341 M * 17.5 mL = 5.9675 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.326 M * 0 mL = 0 mmol

We have:

mol(HI) = 5.968 mmol

mol(NaOH) = 0 mmol

0 mmol of both will react

remaining mol of HI = 5.968 mmol

Total volume = 17.5 mL

[H+]= mol of acid remaining / volume

[H+] = 5.968 mmol/17.5 mL

= 0.341 M

use:

pH = -log [H+]

= -log (0.341)

= 0.4672

Answer: 0.467

2)when 9.15 mL of NaOH is added

Given:

M(HI) = 0.341 M

V(HI) = 17.5 mL

M(NaOH) = 0.326 M

V(NaOH) = 9.15 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.341 M * 17.5 mL = 5.9675 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.326 M * 9.15 mL = 2.9829 mmol

We have:

mol(HI) = 5.968 mmol

mol(NaOH) = 2.983 mmol

2.983 mmol of both will react

remaining mol of HI = 2.985 mmol

Total volume = 26.65 mL

[H+]= mol of acid remaining / volume

[H+] = 2.985 mmol/26.65 mL

= 0.112 M

use:

pH = -log [H+]

= -log (0.112)

= 0.9508

Answer: 0.951

3)

Since this is titration of strong acid and strong base, the solution would be neutral at equivalence point.

So, the pH would be 7.00

Answer: 7.00

4)when 22.5 mL of NaOH is added

Given:

M(HI) = 0.341 M

V(HI) = 17.5 mL

M(NaOH) = 0.326 M

V(NaOH) = 22.5 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.341 M * 17.5 mL = 5.9675 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.326 M * 22.5 mL = 7.335 mmol

We have:

mol(HI) = 5.968 mmol

mol(NaOH) = 7.335 mmol

5.968 mmol of both will react

remaining mol of NaOH = 1.367 mmol

Total volume = 40.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.367 mmol/40.0 mL

= 3.419*10^-2 M

use:

pOH = -log [OH-]

= -log (3.419*10^-2)

= 1.4661

use:

PH = 14 - pOH

= 14 - 1.4661

= 12.5339

Answer: 12.53