A lab technician tests a 0.100 mol/L solution of propanoic acid and finds that its hydrogen ion concentration is 1.16 x 10^-3 mol/L. Calculate the percent ionization of propanoic acid in water
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.
To calculate the percent ionization of propanoic acid in water, we need to first understand the concept of ionization and the relationship between ionization and hydrogen ion concentration.
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%.