Question

Suppose you were given a weak acid with a known pKa of 3.9. You measure a solution of this weak acid and determine it has a pH of 3.6 and the solution contains 50 mM of the non-dissociated conjugate acid (HA). 
What is the concentration of the conjugate base (A-) in the solution?


Answer 1

Data given : pKa = 3.9

                   pH =3.6

                  Concentration of HA, [HA] =50 *10^-3 M

Now we will write the equation,

      [HA]    --------->     [H⁺]        + [A^-]

Ka expression for the above reaction will be ,

Now we will convert pKa value to Ka value

pKa = - log([Ka])

Ka = 10^-pKa

     = 10^-3.9

Ka = 1.2589 *10^-4

Similarly converting pH value to [H⁺]

pH = - log ( [H⁺] )

[H⁺] =10^-pH

       = 10^-3.6

[H⁺] = 2.512 * 10^-4 M

Substituting the values of [H⁺] , Ka, and [HA] in value in the Ka equation

On solving the above equation for [A-] , we get, [A-] = 0.02506 M

Hence, concentration of conjugate base (A- ) in the solution is 0.02506 M