# 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.

###### 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):

1) pH = 0.903

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

2) Recall:

• VNaOH < VHNO3

moles HNO3 = 0.0025 mol HNO3

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

7.14 x 10-3 M

:

2) pH = 2.15

97% (170 ratings) ###### 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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Our tutors have indicated that to solve this problem you will need to apply the Strong Acid Strong Base Titrations concept. You can view video lessons to learn Strong Acid Strong Base Titrations. Or if you need more Strong Acid Strong Base Titrations practice, you can also practice Strong Acid Strong Base Titrations practice problems.

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