Question

Consider a 1 m HNO₂ (Ka = 4.5 x 10⁻⁴) solution. Part a: What is the concentration of H (in m) at equiLiBrium? Round your answer to three places past the decimal. [H] = _______ m. Part b: What is the percent ionization (in

Answer 1

To find the concentration of H+ at equilibrium for a 1 M HNO2 solution, set up an ICE table and solve the quadratic equation derived from the acid dissociation constant (Ka). The equilibrium concentration of H+ is approximately 6.708 x 10^-3 M. Calculate the percent ionization by dividing the equilibrium concentration by the initial concentration and multiplying by 100%.

Step-by-step explanation:

To determine the concentration of hydrogen ions (H+ which is the same as hydronium ions, H3O+) at equilibrium for a 1 M solution of HNO2 with a Ka of 4.5 x 10-4, we can set up an ICE table (Initial, Change, Equilibrium). Assuming that x is the concentration of H3O+ and NO2- at equilibrium and the change for HNO2 is -x, the equilibrium expression based on the acid dissociation constant (Ka) would be:

Ka = [H+][NO2-] / [HNO2] = (x)(x) / (1-x) ≈ x2

Solving the quadratic equation for x when Ka = 4.5 x 10-4 will give us the concentration of H+ at equilibrium. Assuming x << 1, which is reasonable for weak acids, we can simplify to x2 ≈ Ka and find that x, or [H+], is approximately √Ka. Thus, the concentration of H+ is about 6.708 x 10-3 M when rounded to three decimal places.

To calculate the percent ionization, we use the formula:

Percent Ionization = ([H+] / initial concentration) x 100%

Plugging in the calculated [H+] and the initial concentration (1 M) into this formula gives us the percent ionization of the solution.