Question

The Ka of a monoprotic weak acid is 4.98 × 10-3. What is the percent ionization of a 0.191 M solution of this acid?

Answer 1

Answer – We are given, weak monoprotic acid [HA] = 0.191 M ,

Ka = 4.98*10^-3

Need to put ICE table

    HA +   H₂O ------> H₃O⁺ + A^-

I 0.191                    0           0

C -x                        +x         +x

E 0.191-x                 +x          +x

Ka = [H₃O⁺] [A^-] / [HA]

4.98*10^-3 = x*x /(0.191-x)

4.98*10^-3 *(0.191-x) = x²

Now we need to set up quadratic equation

9.51*10^-4 - 4.98*10^-3 x = x²

a =1 , b = 4.98*10^-3, c = -9.51*10^-3

Using the quadratic equation

x = -b +/- √b²-4a*c / 2a

Plugging the value in this formula

x = 0.0951 M

so, x = [H₃O⁺] = 0.0951 M

we know, percent ionization = x / initial concentration *100 %

                                             = 0.0951 / 0.191 *100 %

                                            = 49.8 %