A 25.0 mL sample of 0.150 M nitrous acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The K_aof nitrous acid is 4.50 × 10^-⁴. A) 3.35 B) 7.00 C) 8.11 D) 10.35 E) 10.65

Question

Quizplus · Accepted Answer

At the equivalence point, the number of moles of NaOH added will be equal to the number of moles of HNO2 in the original solution. 
moles of NaOH = 0.150 M x 25.0 mL = 0.00375 mol
moles of HNO2 = 0.150 M x 25.0 mL = 0.00375 mol
The reaction between NaOH and HNO2 is:
HNO2 + NaOH → NaNO2 + H2O
At the equivalence point, all HNO2 is converted to NaNO2. NaNO2 is a salt of a weak base and a strong acid, and it undergoes hydrolysis in water. The reaction is:
NaNO2 + H2O → HNO2 + NaOH
The hydrolysis of NaNO2 gives OH- ions, and the pH increases. 
The pH of the solution at the equivalence point can be calculated using the equation:
pH = pKb + log([salt]/[acid])
where pKb is the negative logarithm of the base dissociation constant of the conjugate base (NO2-), and [salt]/[acid] is the ratio of the concentrations of NaNO2 and HNO2, respectively. 
The pKb of NO2- can be calculated from the pKa of HNO2 as follows:
pKw = pKa + pKb
14.00 = 3.30 + pKb
pKb = 10.70
At the equivalence point, [salt]/[acid] = 1, so 
pH = pKb + log([salt]/[acid]) = pKb = 10.70
However, this is the pH of NaNO2 in water, which is not the same as the pH of the original solution because of dilution. 
Before the titration, the HNO2 solution has a pH of:
pH = pKa + log([NO2-]/[HNO2])
pKa = 3.30
[NO2-]/[HNO2] = 0, because there is no NO2- before the titration
pH = pKa = 3.30
At the equivalence point, the volume of the solution doubles to 50.0 mL, and the number of moles of HNO2 is still 0.00375 mol. Therefore, the concentration of HNO2 is now:
[HNO2] = 0.00375 mol / 0.050 L = 0.075 M
The concentration of OH- can be calculated from the expression for the base dissociation constant of water:
Kw = [H+][OH-]
1.0 x 10^-14 = (x)(x)
x = 1.0 x 10^-7 M
The concentration of OH- from the hydrolysis of NaNO2 is the same as the concentration of NaNO2, which is also 0.075 M.
The total concentration of OH- is 0.075 M + 1.0 x 10^-7 M = 0.075 M (to 3 significant figures). The pH can be calculated from the expression:
pH = 14.00 - log([H+])
[H+] = Kw / [OH-] = 1.33 x 10^-13 M
pH = 12.88 (to 3 significant figures)
However, this is the pH of the solution after excess NaOH has been added. The pH at the equivalence point is between the initial pH of 3.30 and the final pH of 12.88. Since NaNO2 is a basic salt, the pH at the equivalence point is greater than 7.00. The only choice greater than 7.00 is C, which is the correct answer.

A 250 ML Sample of 0