Question

16.6

A 50.0-mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M nitric acid.

You may want to reference (Pages 662 - 674) Section 16.4 while completing this problem.

Part A

Calculate the pH of the solution, after you add a total of 51.9 mL 0.200 M HNO3.

Express your answer using two decimal places.

pH =

nothing

Answer 1

The balanced equation is
       HNO3 (aq) + NaOH (aq) -----------> NaNO3 (aq) + H2O (l)

Number of moles of NaOH = M*V = 0.200 M * 50.0 mL = 10.00 mmol

Number of moles of HNO3 added = 0.200 M * 51.9 mL = 10.38 mmol

This is after equivalence point. At this point, number of moles of HNO3 present in excess.

Number of moles of HNO3 present in excess = 10.38 – 10.00 = 0.38 mmol
Total volume of the solution = 50.0 +51.9 = 101.9 mL

Net conc. of HNO3 = [H+] = 0.38 mmol/101.9 mL = 0.003729 M

pH = – log[H+] = – log(0.003729) = 2.428391 = 2.43

   pH of the solution = 2.43