1)when 0.0 mL of KOH is added
Given:
M(HNO3) = 0.17 M
V(HNO3) = 23.5 mL
M(KOH) = 0.109 M
V(KOH) = 0 mL
mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.17 M * 23.5 mL = 3.995 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.109 M * 0 mL = 0 mmol
We have:
mol(HNO3) = 3.995 mmol
mol(KOH) = 0 mmol
0 mmol of both will react
remaining mol of HNO3 = 3.995 mmol
Total volume = 23.5 mL
[H+]= mol of acid remaining / volume
[H+] = 3.995 mmol/23.5 mL
= 0.17 M
use:
pH = -log [H+]
= -log (0.17)
= 0.7696
Answer: 0.7696
2)when 18.3 mL of KOH is added
Given:
M(HNO3) = 0.17 M
V(HNO3) = 23.5 mL
M(KOH) = 0.109 M
V(KOH) = 18.3 mL
mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.17 M * 23.5 mL = 3.995 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.109 M * 18.3 mL = 1.9947 mmol
We have:
mol(HNO3) = 3.995 mmol
mol(KOH) = 1.995 mmol
1.995 mmol of both will react
remaining mol of HNO3 = 2 mmol
Total volume = 41.8 mL
[H+]= mol of acid remaining / volume
[H+] = 2 mmol/41.8 mL
= 4.785*10^-2 M
use:
pH = -log [H+]
= -log (4.785*10^-2)
= 1.3201
Answer: 1.32
3)
This is titration of strong acid and strong base.
At equivalence point, solution would be neutral as equal mol of acid and base would have been added.
So, pH would be 7.00
Answer: 7.00
4)when 44.9 mL of KOH is added
Given:
M(HNO3) = 0.17 M
V(HNO3) = 23.5 mL
M(KOH) = 0.109 M
V(KOH) = 44.9 mL
mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.17 M * 23.5 mL = 3.995 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.109 M * 44.9 mL = 4.8941 mmol
We have:
mol(HNO3) = 3.995 mmol
mol(KOH) = 4.894 mmol
3.995 mmol of both will react
remaining mol of KOH = 0.8991 mmol
Total volume = 68.4 mL
[OH-]= mol of base remaining / volume
[OH-] = 0.8991 mmol/68.4 mL
= 1.314*10^-2 M
use:
pOH = -log [OH-]
= -log (1.314*10^-2)
= 1.8812
use:
PH = 14 - pOH
= 14 - 1.8812
= 12.1188
Answer: 12.12
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