A 25.0 mL sample of 0.50 M HNO2 is titrated with 0.25 M KOH. What is the ratio of conjugate base to weak acid, at the point where 10.0 mL of KOH has been added during the titration? Ka for nitrous acid is 5.6 x 10-4. a. 5.0 b. 0.20 c. 0.25 d. 2.0

the ratio of conjugate base to acid is 0.25

Explanation:

The titration reaction is

HNO₂ + KOH   ⇒ K⁺ + NO₂⁻ + H₂O

where NO₂⁻ is the conjugate base of the weak acid HNO₂

Since we are being asked to find the ratio of conjugate base to weak acid we know that we are going to have some HNO₂ unreacted after the addition of 10.0 mL (0.010 L) KOH.

We have all the information required to do the calculation from the stoichiometry of the reaction.

mol KOH = M x V

where M is the molarity, and V is the volume in liters:

mol KOH = 0.25 M x 0.010 L = 0.25 mol/L x 0.010 L

                                               = 0.0025 mol

mol KNO₂ produced = 0.0025 mol

mol HNO₂ before reaction = 0.50 M x 0.025 L = 0.0125 mol

mol HNO₂  after reaction = 0.0125 mol - 0.0025 mol = 0.010 mol

Therefore the mol ratio will be

NO₂⁻ / HNO₂  = 0.0025/0.010 = 0.25

Incidentally this will be the same ratio in M, since the volumes will cancel out in the ratio.

answersilicon (si); oxygen (o);