What is the pH at the equivalence point?

Question

70.0 mL of 0.250M HNO2 is titrated to its equivalence point with 1.00M NaOH. What is the pH at the equivalence point?

Answer 1

HNO₂ + NaOH ==> NaNO₂ + H₂O ... balanced equation

At the equivalence point, the moles of HNO₂ = moles NaOH and so all of the HNO₂ and NaOH are converted to NaNO₂ and H₂O. The pH of the solution is then determined by the NaNO₂. Since it is the salt of a weak acid and a strong base, the pH will be basic (>7). To find the pH, we find the [NaNO₂] and then look at the hydrolysis.

Concentration of NO₂^-:

moles NO₂^- = 70.0 ml x 1 L / 1000 ml x 0.250 mol / L = 0.0175 moles

volume OH- = 0.0175 mols OH- x 1 L / 1.00 mol/L = 0.0175 L

Final volume of solution = 0.070 L + 0.0175 L = 0.0875 L

Concentration of NO₂^- = 0.0175 mols NO₂^-- / 0.0875 L = 0.200 M NO₂^-

Hydrolysis and pH of NO2- solution:

NO₂^- + H2O ==> HNO₂ + OH^-

Here, NO₂^- is acting as a base, so we need the Kb for NO₂^--. We can look it up, or look up the Ka of HNO₂ and then find the Kb using Kb = 1x10^-14 / Ka. I find the Ka for HNO₂ to be 4.0x10^-4. Thus,

Kb = 1x10^-14/4x10^-4 = 2.5x10^-11

Kb = [HNO₂][OH^-] / [NO₂^-]

2.5x10^-11 = (x)(x) / 0.200

x² = 5x10^-12

x = 2.23x10^-6 M = [OH-]

pOH = -log 2.23x10^-6 = 5.65

pH = 14 - pOH

pH = 8.35 (basic as originally predicted)