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

Question

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

Calculate pH:

after adding 30.00 mL of HNO3

at the equivalence point

Answer 1

Sodium hydroxide (NaOH) is a strong base, and nitric acid (HNO3) is a strong acid.

NaOH + HNO₃ ==> NaNO₃ + H₂O ... balanced equation

Initial moles of NaOH = 50.0 ml x 1 L/1000 ml x 0.200 mols/L = 0.0100 moles NaOH

Moles of HCl in 30.00 ml of 0.200 M HNO₃ = 30.00 ml x 1 L/1000 ml x 0.200 mol/L = 0.00600 mols HNO₃

Set up an ICE table:

NaOH + HNO₃ ==> NaNO₃ + H₂O

0.0100.....0.00600......0.........................Initial

-0.006......-0.006.......+0.006.................Change

0.004...........0.............0.006.................Equilibrium

Final volume after addition = 50 ml + 30 ml = 80 mls = 0.0800 L

Final [NaOH] = 0.004 mols / 0.0800 L = 0.05 M

pOH = -log [OH-] = -log 0.05 = 1.301

pH = 14 - 1.301

pH = 12.699 (3 sig.figs.)

At the equivalence point, the moles of NaOH equals the moles of HNO₃, so the only species are NaNO₃ and H₂O. Since this is a reaction between a strong acid and a strong base, the pH at equivalence will be pH = 7.