Question

The Ka of hydrazoic acid (HN3) is 1.9 × 10-5 at 25.0 °C. What is the pH of a 0.40 M aqueous solution of HN3?

Answer 1

Answer:

The pH of the solution is 2.56.

Explanation:

Given :

Concentration (c) = 0.40 M

Acid dissociation constant = $K_a=1.9\times 10^{-5}$

The equilibrium reaction for dissociation of $HN_3$ (weak acid) is,

                           $HN_3+H_2O\rightleftharpoons N_3^{-}+H_3O^+$

initially conc.         c                       0         0

At eqm.              $c(1-\alpha)$                $c\alpha$        $c\alpha$

Dissociation constant is given as:

$K_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

$1.9\times 10^{-5}=\frac{(0.40\alpha)(0.40\alpha)}{0.40(1-\alpha)}$

By solving the terms, we get value of $(\alpha)}$

$\alpha=0.00686832$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.4\times 0.00686832=0.002747 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.002747)$

$pH=2.56$

Therefore, the pH of the solution is, 2.56.