Question

What is the percent dissociation of HX (pKa = 4.30) in a 0.380 M solution?

________ %

Answer 1

Answer – Given, [HX] = 0.380 M , pKa = 4.30

Now first we need to calculate the Ka from pKa

We know,

pKa = -log Ka

so, Ka = 10^-pKa

            = 10^-4.30

            = 5.01*10^-5

Need to put ICE table for calculating the [H⁺] or [H₃O⁺]

    Hx +   H₂O ------> H₃O⁺ + x^-

I 0.380                   0           0

C -x                    +x         +x

E 0.380-x            +x          +x

Ka = [H₃O⁺] [A^-] / [HA]

5.01*10^-5 = x*x /(0.380-x)

We can neglect the x in the 0.380-xm, since Ka value is too small

7.30*10^-3 *0.380 = x²

x = 0.00436 M

so, x = [H₃O⁺] = 0.00436 M

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

                                             = 0.00436 / 0.380 *100 %

                                            = 1.15 %