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_a of nitrous acid is 4.6 × 10^-4. A) 10.35 mL B) 10.65 mL C) 3.35 mL D) 7.00 mL E) 8.11 mL

Question

Quizplus · Accepted Answer

Since nitrous acid is a weak acid, the equivalence point will have a pH greater than 7 due to the presence of excess hydroxide ions. To calculate the volume of NaOH required to reach the equivalence point, we can use the equation:

M1V1 = M2V2

where M1 and V1 are the concentration and volume of nitrous acid, and M2 and V2 are the concentration and volume of NaOH at the equivalence point. Solving for V2 gives:

V2 = (M1V1) / M2
   = (0.150 M)(25.0 mL) / 0.150 M
   = 25.0 mL

Since the volumes of acid and base are equal at the equivalence point, the total volume of the solution is 50.0 mL. To calculate the concentration of nitrite ions (the conjugate base of nitrous acid) at the equivalence point, we can use the fact that nitrous acid and nitrite ions are in a 1:1 ratio:

moles of nitrite ions = moles of nitrous acid at equivalence point
                      = (0.150 M)(25.0 mL) / 1000 mL/L
                      = 0.00375 mol

concentration of nitrite ions at equivalence point = moles / volume
                                                  = 0.00375 mol / 50.0 mL
                                                  = 0.075 M

To calculate the pH at the equivalence point, we can use the fact that nitrite ions undergo hydrolysis in water and produce hydroxide ions:

NO2- + H2O ⇌ HNO2 + OH-

The equilibrium constant for this reaction is:

Kb = Kw / Ka
   = 1.0 × 10^-14 / 4.6 × 10^-4
   = 2.17 × 10^-11

Since the NaOH solution is also 0.150 M, we can assume that all of the hydroxide ions come from the NaOH and use the equation for Kb:

Kb = [NO2-][OH-] / [HNO2]

0.075 M × [OH-] / 4.6 × 10^-4 = 2.17 × 10^-11

[OH-] = 1.83 × 10^-7 M

pOH = -log[OH-] = 6.74

pH = 14.00 - pOH = 7.26

Therefore, the pH at the equivalence point is 7.26. The closest answer choice is E, 8.11 mL, which assumes a slight excess of NaOH. However, the actual equivalence point occurs at exactly 25.0 mL, when the volumes of acid and base are equal.

A 250 ML Sample of 0