titration of 25.0 mL of 0.150 M HNO3 with 0.150 M NaOH, perform the calculations
ID: 509803 • Letter: T
Question
titration of 25.0 mL of 0.150 M HNO3 with 0.150 M NaOH, perform the calculations required to check agreement with experimental values for the following volumes of titrant.
(For points before the equivalence point, the calculations required are amount acid reacted, amount acid remaining, concentration of acid remaining, and pH.
For points after the equivalence point, the calculations required are amount of excess hydroxide, concentration of hydroxide, pOH, and pH.)
1. 12.0 mL
2. 22.5 mL
3. 38.0 mL
Explanation / Answer
HNO3 + NaOH -------------> NaNO3 + H2O
millimoles of HNO3 = 25 x 0.15 = 3.75
1) 12 mL of NaOH added
millimoles NaOH added = 12 x 0.15 = 1.8
3.75 - 1.8 =1.95 millimoles HNO3 left
total volume = 25 + 12 = 37mL
[HNO3] = 1.95 / 37 = 0.0527 M HNO3 left
pH = - log [H+]
pH = - log [0.0527]
pH = 1.28
2) 22.5 mL NaOH added
millimoles of NaOH = 22.5 x 0.15 = 3.375
3.75 - 3.375 = 0.375 millimoles HNO3 left
total volume = 25 + 22.5 = 47.5 mL
[HNO3] = 0.375 / 47.5 = 0.00789 M
pH = - log [0.00789]
pH = 2.10
3) 38 mL NaOH added
millimoles of NaOH added = 38 x 0.15 = 5.7
5.7 - 3.75 = 1.95 millimoles NaOH exess present
total volume = 25 + 38 = 63 mL
[NaOH] = 1.95 / 63 = 0.031 M
pOH = - log [OH-]
pOH = - log [0.031]
pOH = 1.51
pH = 14 - 1.51
pH = 12.49
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