Question

a)What is the percent ionization of a 0.0682 M aqueous solution of acetic acid? K_a (CH₃COOH) = 1.8x10^-5

^b)What is the percent ionization of a 0.472 M aqueous solution of formic acid? Ka (HCOOH) = 1.7x10^-4

^c)What is the percent ionization of a 0.419 M aqueous solution of carbonic acid? K_a1 = 4.2x10^-7; K_a2 = 4.8x10^-11

^d)What is the percent ionization of a 0.493 M aqueous solution of hydrosulfuric acid? K_a1 = 9.5x10^-8; K_a2 = 1x10^-19

Answer 1

a)

Since acetic acid is a weak acid in water, the degree of ionization to H+ and CH3COO- is very small implying the concentration of CH3COOH is 0.0682M.

let the concentration of [H+] = [CH3COO-] =x

then, [H+][CH3COO-]/[CH3COOH] = Ka,

x^2/0.0682 = 1.8X10-5

x= 1.1X10-3 M = [H+]

then, % ionization = [H+]/[HA]*100 = 1.1x10-3/0.0682*100 = 1.6%

b)

In this part, let there be x M of [H+] and [HCOO-] at equilibrium, then there will be 0.472-x M [HCOOH], implying

Ka = [H+][HCOO-]/[HCOOH]

1.7x10-4 = x^2/(0.472-x), Solving for x we get, x= 0.009 M

then % ionization = [H+]/[HA]*100 = 0.009/0.472*100 = 1.9%

c)

Since carbonic acid is a diprotic acid, considering the first ionization

Ka1 = [H+][HCO3-]/[H2CO3] ; at equilibrium if [H+]= x M, then [HCO3-]=x M and [H2CO3] = 0.419-x M

4.2x10-7 = x^2/(0.419-x) , assuming x<<<0.419 as Ka is very small, would give us 4.2x10-7 = x^2/0.419, then x= 4.19x10-4 M

Since Ka2 is even smaller than Ka1, and initial concentration of [HCO3-] is 4.19x10-4 M, the [H+] generated from second ionization will be negligible.

implying that the percent ionization would be calculated from first ionization only which will be, [H+]/[H2CO3]*100 = 4.19x10-4/0.419*100 = 1%.