1. When NO2 is bubbled into water, it is completely converted to HNO3 and HNO2 a

Question

1. When NO2 is bubbled into water, it is completely converted to HNO3 and HNO2 according to the reaction shown below. Calculate the pH and the molar concentrations of all dissolved species (H3O+, OH–, HNO2, HNO3, NO2–, and NO3–) in a solution prepared by bubbling 4.601 g NO2 through 2.00 L of water.

2NO2(g) + H2O(l) HNO3(aq) + HNO2(aq)

2. 12.6-g sample of Na2A (molar mass = 126 g/mol) is dissolved enough water to produce 1.00 L of solution. The pH of this solution is 9.85 and the equilibrium concentration of H2A is 1.0 × 10–12 M. Determine the values of pKa1 and pKa2 for H2A

Explanation / Answer

The solutions are as shown below

1. Use Ka value from data table

Ka = [HNO3][HNO2]/[NO2]^2

Calculate molar concentration of NO2 = 4.601 g / 46 x 2 = 0.05 M

let x be the change i concentration,

Ka = (x)^2/(0.05-x)^2

Solve for x,

x = [HNO3] = [HNO2]

[NO2] = 0.05 - x

2. calculate molar concentration of Na2A

= 12.6 g / 126 g/mol x 1 L = 0.1 M

pH = 9.85

[H+] = 1.41 x 10^-10

Ka = [H+][HA-]/[H2A]

     = (1.41 x 10^-10)^2/(1 x 10^-12)

     = 2 x 10^-19

pKa1 = 18.7

let x be the change in concentration at equilibrium then we have,

pH = pKa + log[base]/[acid]

9.85 = 18.7 + log(x)/1 x 10^-12)

x = [Na2A] = 1.4 x 10^-21

pKa2 = 20.9

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.