Question

Determine the equilibrium constant for the acid-base reaction between ethanol and hydrobromic acid? Acid pKa Hydrobromic Acid −5.8 Ethyloxonium Ion −2.4

Answer 1

Answer:

$10^{-3.4$

Explanation:

Let us first take a look at the image below;

In the acid - base  reaction; we can see the transfer of electrons that takes place;

We can also see that the reaction goes in the direction which converts the stronger acid and the stronger base to the weaker acid and the weaker base.

The stronger acid is shown with the one with more negative $P_{Ka}$ Value.

∴ The equilibrium constant for the acid-base reaction is expressed as:

$K_{eq}= \frac{K_a of reactant acid}{K_a of product acid}$

      $= \frac{10^{-pK} of reactant acid}{10^{-pk} of product acid}$

From  $P_{Ka}$ Value (shown in the image below), it is clear and vivid that hydrobromic acid is a stronger acid than the ethyloxonium ion, therefore the equilibrium lies to the right.

From the chemical equation (shown in the attached image); the equilibrium constant for the acid-base reaction can be expressed as:

$K_{eq}=\frac{10^{-pK} of hydrobromic acid}{10^{-pk} of Ethyloxonium acid}$

$K_{eq}=\frac{10^{-5.8} of hydrobromic acid}{10^{-2.4} of Ethyloxonium acid}$

       $= 10^{-3.4}$