the Ka of a monoprotic weak acid is 0.00247. what is the percent ionization of a 0.157 m solution of this acid?

The ratio of the equilibrium H3O+ concentration to the initial HA concentration, multiplied by 100%, is the definition of the weak acid, HA, percent ionization.

Ionization energy, sometimes referred to as ionisation potential, is the amount of energy needed to remove an electron from an isolated atom or molecule in physics and chemistry.

In an electric discharge tube, when a quickly moving electron produced by an electric current collides with a gaseous atom of the element, causing it to eject one of its electrons, the ionisation energy of a chemical element, given in joules or electron volts, is often measured.

You must have been referring to this monoprotic acid, which we will refer to as HA, since the Ka value is 0.00584.

HA <==> H+ + A-

Ka = 0.00584 = [H+]

[A-]/[HA]

Assuming that x is modest in comparison to 0.181, 0.00584 = (x)(x)/0.181 - x, we shall disregard it for the time being.

x2 = 0.001057

Since our assumption was erroneous and this value is closer to 18% of 0.180, we must now utilize the quadratic equation to solve for x. x = [H+] = [A-] = 0.0325

x2 = 0.001057 - 0.00584x

x = 0.0297

Ionization percentage equals 0.0297/0.181 (100%) = 16.4%.

Learn more about Percentage Ionization from given link

https://brainly.in/question/1695326

#SPJ4