Question

A certain weak acid has a dissociation constant of 1.0104, the equilibrium constant for its reaction with the strongest base is:

Answer 1

Answer:

Keq = 10⁻¹⁰

Explanation:

We Have :-

Dissociation Constant = 1 * 10⁻⁴

To Find :-

Equilibrium Constant

Solution :-

HA ⇆ H⁺ + A⁻

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

And

HA + B⁺ + OH ⇆ B⁺ + A⁻ + H₂O

$Keq = \frac{[H2O][A^{-}] }{[HA][OH]} \\\\\frac{Keq}{Ka} = \frac{1}{[H^{+}][OH^{-}] } \\\\\frac{Keq}{Ka} = \frac{1}{Kwater}$

$Keq = \frac{Ka}{Kwater} \\\\= \frac{10^{-4} }{10^{-14} } \\\\= 10^{-10}$

Answer 2

Answer : 10^10

Solution is in attachment