Question

A 0.100 m solution of a monoprotic weak acid has a ph of 3.00. what is the pka of this acid?

Answer 1

Explanation:

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

             pH = $-log [H^{+}]$

                  3 = $-log [H^{+}]$

                  antilog (-3) = $1 \times 10^{-3}$

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

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

           $HA \rightarrow H^{+} + A^{-}$

                  0.1           0         0

                0.1 - x         x         x

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

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

                    = $\frac{x \times x}{(c - x)}$

                     = $\frac{1 \times 10^{-3} \times 1 \times 10^{-3}}{(0.1 - 1 \times 10^{-3})}$

                  = $1.01 \times 10^{-5}$

Also,     $pK_{a} = -log K_{a}$

                        = $-log (1.01 \times 10^{-5})$

                       = 5.00

Thus, we can conclude that the $pK_{a}$ of given acid is 5.0.

Answer 2

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.