Let's call propanoic acid HP.
HP ==> H^+ + P^-
Ka = (H^+)(P^-)/(HP)
If 0.1 M HP acutaly has 1.16 x 10^-3 (H^+), then
ionization = 1.16 x 10^-3/0.1 and you can convert that to percent by multiplying by 100.
HP ==> H^+ + P^-
Ka = (H^+)(P^-)/(HP)
If 0.1 M HP acutaly has 1.16 x 10^-3 (H^+), then
ionization = 1.16 x 10^-3/0.1 and you can convert that to percent by multiplying by 100.
When a substance like propanoic acid (CH3CH2COOH) dissolves in water, it undergoes ionization and forms hydrogen ions (H+) and the corresponding conjugate base ions (CH3CH2COO-). The extent of ionization can be quantified by the concentration of hydrogen ions in solution.
The percent ionization can be calculated using the following formula:
Percent Ionization = (Hydrogen Ion Concentration / Initial Acid Concentration) * 100
In this case, the hydrogen ion concentration is given as 1.16 x 10^-3 mol/L, and the initial acid concentration is 0.100 mol/L.
Plugging in the given values:
Percent Ionization = (1.16 x 10^-3 mol/L / 0.100 mol/L) * 100
Calculating the value:
Percent Ionization = (0.0116) * 100
Percent Ionization ≈ 1.16%
Therefore, the percent ionization of propanoic acid in water is approximately 1.16%.