00:02
Problem 64 part a.
00:10
In question, given that propionic acid dissociated into hydrogen ion and propionate ion, in part a, initial concentration of propionic acid is 0 .250.
00:30
So at equilibrium, concentration of propionic acid is 0 .250.
00:42
So at equilibrium, concentration of propionic acid, let x small of hydrogen and is formed at equilibrium therefore x small of propionate and is also formed equilibrium therefore remaining concentration of propionic acid as equilibrium is 0 .250 minus x m okay now we can write k a for this reaction concentration of reactant divided by concentration of product so hydrogen ion concentration multiplied by propionate ion divided by concentration of propionic acid that is equals to x squared divided by 0 .250 minus x again this value of x is very small as compared to 0 .250 so we can write as x is squared divided by 0 .250 and value of k according to appendix d is for propionic acid is 1 .3 into 10 to the power minus 5 now we can after calculating we can write value of x equals to 1 .803 multiply by 10 to the power minus 3 m, hydrogen ion concentration.
02:31
Now, percentage ionization equals to we know h -in concentration divided by initial acid concentration multiplied by 100.
02:50
This h2ndon concentration at equilibrium.
02:53
So put the value 1 .803 into 10 to the power minus 3m, hydrogen ion, divided by initial concentration of acid is 0 .250 multiplied by 100...