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
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
Determine the pH during the titration of 22.6 mL of 0.185 M nitric acid by 0.294...
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