Determine the pH at the equivalence (stoichiometric) point in the titration of 40 mL

Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 40 mL of 0.12 M HNO₂(aq) ith 0.1 M NaOH(aq). The K_a of HNO₂ is 7.1 x 10^-4.

The answer is 7.94

Answer 1

Since HNO2 is a weak acid, and you are titrating it with a strong base, you will be forming a buffer because you will have a weak acid and the salt of that weak acid.

HNO₂ + OH^- ====> NO₂^- + H₂O

weak acid................. salt..........water

At equivalence moles OH = moles of HNO2

moles HNO₂ = 40 ml x 1 L/1000 ml x 0.12 mol/L = 0.0048 moles HNO₂

Since we need to add 0.0048 moles NaOH, we can find the volume of NaOH needed:

(0.1 mol/L)(x L) = 0.0048 moles

x = 0.048 liters = 48 mls

TOTAL VOLUME = 40 ml HNO₂ + 48 mls NaOH = 88 mls = 0.088 liters

Final [NO₂^-] = 0.0048 moles/0.088 L = 0.055 M

At equivalence, all of the HNO₂ has been converted to NO₂^- (as NaNO₂), and the NO₂^- will hydrolyze....

NO₂^- + H₂O ==> HNO₂ + OH^- (here the NO₂^- is acting as a base so we will use the K_b)

K_b = Kw/Ka = 1x10^-14/7.1x10^-4 = 1.4x10^-11

K_b = [HNO₂][OH^-]/[NO₂^-]

1.4x10^-11 = (x)(x)/0.055

x2 = 7.6x10^-13

x = 8.7x10^-7 = [OH^-]

pOH = 6.06

pH = 7.94