Question

A 0.50 M solution of an unknown acid has a pH = 4.0. Of the following, which is the acid in the solution?


HOCl (Ka = 2.0 x 10-8)
HBr (strong acid)
HF (Ka = 6.8 x 10-4)
C6H5OH (Ka = 1.0 x 10-10)

Answer 1

Answer:

$HOCl$, $K_a=2.0\times 10^{-8}$

Explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,  

pH = - log [H⁺]

Strong acids dissociate completely and thus, 0.5 M of a solution of a strong acid yields a pH of 1.0 .

The expression of the pH of the calculation of weak acid is:-

$pH=-log(\sqrt{k_a\times C})$

Where, C is the concentration = 0.5 M

Given, pH = 4.0

So, for $HOCl$, $K_a=2.0\times 10^{-8}$

$pH=-log(\sqrt{2.0\times 10^{-8}\times 0.5})$

pH = 4.0

Hence, the acid is HOCl.