Calculate the percent ionization of nitrous acid in a solution that is 0.311 M in nitrous acid (HNO2) and 0.189 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.50 × 10-4.

To calculate the percent ionization of nitrous acid (HNO2) in the given solution, we can use the acid dissociation constant (Ka) and the initial concentration of the acid.

The equation for the ionization of nitrous acid is:

HNO2 ⇌ H+ + NO2-

The acid dissociation constant (Ka) is given as 4.50 × 10^-4.

We are given the initial concentration of nitrous acid (HNO2) as 0.311 M.

Step 1: Write the expression for the percent ionization:

Percent ionization = (concentration of H+ ions / initial concentration of HNO2) × 100

Step 2: Calculate the concentration of H+ ions at equilibrium:

We can assume that initially, x amount of the nitrous acid ionizes, resulting in x amount of H+ ions and x amount of NO2- ions. Therefore, the concentration of H+ ions at equilibrium is also x.

Step 3: Use the acid dissociation constant (Ka) to write an expression for the equilibrium concentration of H+ ions and NO2- ions:

Ka = [H+][NO2-] / [HNO2]

Since the concentration of H+ ions and NO2- ions is x and the initial concentration of HNO2 is 0.311 M, we can substitute these values in the equation:

4.50 × 10^-4 = x^2 / 0.311

Solving for x, we find:

x ≈ 0.00539 M

Step 4: Calculate the percent ionization:

Percent ionization = (concentration of H+ ions / initial concentration of HNO2) × 100
= (0.00539 / 0.311) × 100
≈ 1.73%

Therefore, the percent ionization of nitrous acid (HNO2) in the given solution is approximately 1.73%.

To calculate the percent ionization of nitrous acid in the given solution, we need to use the formula for percent ionization, which is equal to the concentration of the ionized species divided by the initial concentration of the acid, multiplied by 100%.

Here's how you can calculate it step by step:

Step 1: Identify the balanced chemical equation for the ionization of nitrous acid (HNO2):
HNO2(aq) <--> H+(aq) + NO2-(aq)

Step 2: Determine the initial concentration of nitrous acid (HNO2) in the solution, which is given as 0.311 M.

Step 3: Determine the equilibrium concentration of the ionized species (H+ and NO2-) by using the acid dissociation constant (Ka). The Ka for nitrous acid is given as 4.50 × 10^-4. Let's assume x is the concentration of H+ and NO2- at equilibrium.

Ka = [H+][NO2-]/[HNO2]
4.50 × 10^-4 = x^2 / (0.311 - x)

Since the initial concentration of HNO2 (0.311 M) is much larger than x, we can assume that x is small compared to the initial concentration. Thus, we can approximate 0.311 - x as 0.311.

4.50 × 10^-4 = x^2 / (0.311)
x^2 = 4.50 × 10^-4 * 0.311
x^2 ≈ 1.395 × 10^-4
x ≈ √(1.395 × 10^-4)
x ≈ 0.0118 M

Step 4: Calculate the percent ionization of nitrous acid:
Percent Ionization = (concentration of ionized species / initial concentration of HNO2) * 100%
Percent Ionization = (0.0118 M / 0.311 M) * 100%
Percent Ionization ≈ 3.79%

Therefore, the percent ionization of nitrous acid in the given solution is approximately 3.79%.

oidk

........HNO2 ==> H^+ + NO2^-

I.....0.311......0.......0
C......-x.........x........x
E.....0.311-x......x......x

......KNO2 ==>K^+ + NO2^-
I.....0.189...0......0
C....-0.189..0.189...0.189
E......0.....0.189..0.189

Ka = (H^+)(NO2^-)/(HNO2)
Substitue for Ka,
For (H^+) = x
For (NO2^-) = x(from HNO2) + 0.189(from KNO2)
(HNO2) = 0.311-x
Solve for x = (H^+), then
%ion = [(H^+)/(0.311)]*100 = ?