Problem: A 20.0-mL sample of 0.125 M HNO3 is titrated with 0.150 M NaOH. Calculate the pH for at least five different points throughout the titration curve and make a sketch of the curve.

🤓 Based on our data, we think this question is relevant for Professor Harwood's class at UNI.

FREE Expert Solution

Five points:

  1. start of titration
  2. VNaOH < VHNO3
  3. equivalence point: pH = 7
  4. VNaOH = VHNO3
  5. VNaOH > VHNO3


1) At start of titration:


When getting the pH of a strong acid solution (HNO3):

pH=-log[strong acid]pH=-log[HNO3] pH=-log[0.125 M]

1) pH = 0.903


Reaction:            HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)


2) Recall:

molarity (M)=molL

  • VNaOH < VHNO3


moles HNO3=0.125 mol HNO3L 20.0 mL x 10-3 L1 mL

moles HNO3 = 0.0025 mol HNO3

moles NaOH=0.150 mol NaOHL15.0 mL x 10-3 L1 mL

moles NaOH = 0.00225 mol NaOH


moles NaOH < moles HNO3pH depends on molarity of acid (NaOH is consumed)

  • moles HNO3 reacted = moles NaOH
  • moles HNO3 = 0.0025 - 0.00225 = 2.5 x 10-4


[HNO3]f=2.5 ×10-4 mol35.0  mL×10-3 L1 mL=7.14 x 10-3 M

:

pH=-log[7.14×10-3 M]

2) pH = 2.15

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Problem Details

A 20.0-mL sample of 0.125 M HNO3 is titrated with 0.150 M NaOH. Calculate the pH for at least five different points throughout the titration curve and make a sketch of the curve.

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What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Harwood's class at UNI.