Question

In the laboratory a student measures the percent ionization of a 0.510 M solution of hydrofluoric acid to be 3.87 %.

Calculate value of K_a from this experimental data.

K_a =  

Answer 1

Let the acid be HA

HA       <—>   H+   +   A-

0.510       0       0   (initial)

0.510-x       x       x   (at equilibrium)

% ionisation = x*100/c

3.87 = x*100/0.510

x = 0.01974 M

Ka = [H+][A-]/[HA]

Ka = x*x/(0.510-x)

Ka = (0.01974)*(0.01974)/(0.510 - 0.01974)

Ka = 7.95*10^-4

Answer: 7.95*10^-4