Question

A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentrations of the reactants and products are [HA]=0.250 M , [H+]=2.00×10−4 M , and [A−]=2.00×10−4 M . Calculate the value of pKa for the acid HA .

Answer 1

Answer:

$pK_{a}$ of HA is 6.80

Explanation:

$pK_{a}=-logK_{a}$

Acid dissociation constant ($K_{a}$) of HA is represented as-

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

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

$K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}$

So, $K_{a}=1.6\times 10^{-7}$

Hence $pK_{a}=-log(1.6\times 10^{-7})=6.80$