2.0M Propionic Acid HC3H5O2 Dissolves In Distilled Water. If It Has A Ka Of 1.3*10-5, What Is The Final PH? The base ethylamine C2H5NH2 has a Kb of 5.6*10-4. What will the pH be in .53 molar solution?

first you have to calculate the amount ionized. We will say it is x mol / L

then % ionization = (amount ionized / initial concentration) * 100%

x can be calculated using an ice chart

HC3H5O2           > H+    + C3H5O2-

Initial HC3H5O2 = 0.250  

change              = -x

equilibrium         = 0.250 - x

initial H+            = 0

change              = +x

equilibrium         = x

C3H5O2- initial   = 0

change              = +x

equilibrium         = x

Ka         = [H=][C3H5O2-] / HC3H5O2]

 1.3 * 10 ^ -5    = [x][x] / (0.250 - x)  

So 1.3 * 10 ^ -5 * (0.250 - x)   = x ^ 2  

     3.25 * 10^ -6 - (1.3 * 10^-5)x = x^2   now this is a quadratic equation and you have to rearrange it and solve for x

x^2 + 1.3 * 10^-5)x - 3.25 * 10^ -6 = 0

use the equation x = {-b (+ or -)[b^2 - 4.a.c] ^ 1/2} / 2a

you should get x = 1.80 * 10 ^ -3 or x = -1.80* 10^-3

but x can not be negative..

so x = 1.80 * 10 ^ -3

so percent ionization = (1.80 * 10 ^ -3 / 0.250) * 100%

                               =0.72 %

the other way which is more easier is  

assuming that x is very small and therefore 0.250 - x is approximately equals to 0.250  

then 1.3 * 10^-5 = x^2 / 0.250

so x^2    = 1.3 * 10^-5 * 0.250

        x   = 1.80 * 10 ^-3

then percent ionization is = (1.80 * 10 ^ -3 / 0.250) * 100%

                               =0.72 %

if the percent ionization is > 5 % you can not do that approximation. in such a case you have to solve the quadratic equation. that is why I showed both methods.

now you can do the parts b and c

b answer : percent ionization = 1.27 %  

c answer : 2.54%

good luck