Determine the pH during the titration of 14.7
mL of 0.118 M hydroiodic acid by
8.94×10-2 M sodium
hydroxide at the following points:
(1) Before the addition of any sodium
hydroxide
(2) After the addition of 9.70 mL of
sodium hydroxide
(3) At the equivalence point
(4) After adding 24.6 mL of sodium
hydroxide
1)when 0.0 mL of NaOH is added
Given:
M(HI) = 0.118 M
V(HI) = 14.7 mL
M(NaOH) = 0.0894 M
V(NaOH) = 0 mL
mol(HI) = M(HI) * V(HI)
mol(HI) = 0.118 M * 14.7 mL = 1.7346 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.0894 M * 0 mL = 0 mmol
We have:
mol(HI) = 1.735 mmol
mol(NaOH) = 0 mmol
0 mmol of both will react
remaining mol of HI = 1.735 mmol
Total volume = 14.7 mL
[H+]= mol of acid remaining / volume
[H+] = 1.735 mmol/14.7 mL
= 0.118 M
use:
pH = -log [H+]
= -log (0.118)
= 0.9281
Answer: 0.928
2)when 9.7 mL of NaOH is added
Given:
M(HI) = 0.118 M
V(HI) = 14.7 mL
M(NaOH) = 0.0894 M
V(NaOH) = 9.7 mL
mol(HI) = M(HI) * V(HI)
mol(HI) = 0.118 M * 14.7 mL = 1.7346 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.0894 M * 9.7 mL = 0.8672 mmol
We have:
mol(HI) = 1.735 mmol
mol(NaOH) = 0.8672 mmol
0.8672 mmol of both will react
remaining mol of HI = 0.8674 mmol
Total volume = 24.4 mL
[H+]= mol of acid remaining / volume
[H+] = 0.8674 mmol/24.4 mL
= 3.555*10^-2 M
use:
pH = -log [H+]
= -log (3.555*10^-2)
= 1.4492
Answer: 1.45
3)
So, pH at equivalence point will be neutral which is 7.00
Answer: 7.00
4)when 24.6 mL of NaOH is added
Given:
M(HI) = 0.118 M
V(HI) = 14.7 mL
M(NaOH) = 0.0894 M
V(NaOH) = 24.6 mL
mol(HI) = M(HI) * V(HI)
mol(HI) = 0.118 M * 14.7 mL = 1.7346 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.0894 M * 24.6 mL = 2.1992 mmol
We have:
mol(HI) = 1.735 mmol
mol(NaOH) = 2.199 mmol
1.735 mmol of both will react
remaining mol of NaOH = 0.4646 mmol
Total volume = 39.3 mL
[OH-]= mol of base remaining / volume
[OH-] = 0.4646 mmol/39.3 mL
= 1.182*10^-2 M
use:
pOH = -log [OH-]
= -log (1.182*10^-2)
= 1.9273
use:
PH = 14 - pOH
= 14 - 1.9273
= 12.0727
Answer: 12.07
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