Respuesta :

Answer is: pKa for the monoprotic acid is 5.
Chemical reaction: HA(aq) 
⇄ A⁻(aq) + H⁺(aq).
c(monoprotic acid) = 0.100 M.
pH = 3.00.
[A
⁻] = [H⁺] = 10∧(-3).
[A
⁻] = [H⁺] = 0.001 M; equilibrium concentration.
[HA] = 0.1 M - 0.001 M.
[HA] = 0.099 M.
Ka = [A
⁻]·[H⁺] / [HA]. 
Ka = (0.001 M)² / 0.099 M.
Ka = 0.00001 M = 1.0·10
⁻⁵ M.
pKa = -logKa = 4.99.

Explanation:

It is known that the relation between pH and hydrogen ion concentration is as follows.

             pH = [tex]-log [H^{+}][/tex]

                  3 = [tex]-log [H^{+}][/tex]

                  antilog (-3) = [tex]1 \times 10^{-3}[/tex]

                     [tex][H^{+}] = 1 \times 10^{-3}[/tex] M

Now, let the given acid is HA and it dissociates as follows.

           [tex]HA \rightarrow H^{+} + A^{-}[/tex]

                  0.1           0         0

                0.1 - x         x         x

We know that relation between [tex]K_{a}[/tex], [tex][H^{+}][/tex] and HA is as follows.

         [tex]K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}[/tex]

                    = [tex]\frac{x \times x}{(c - x)}[/tex]

                     = [tex]\frac{1 \times 10^{-3} \times 1 \times 10^{-3}}{(0.1 - 1 \times 10^{-3})}[/tex]

                  = [tex]1.01 \times 10^{-5}[/tex]

Also,     [tex]pK_{a} = -log K_{a}[/tex]

                        = [tex]-log (1.01 \times 10^{-5})[/tex]

                       = 5.00

Thus, we can conclude that the [tex]pK_{a}[/tex] of given acid is 5.0.