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 ___
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
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