Question

Determine the pH of a 0.500 M HNO2 solution. Ka of HNO2 is 4.6 * 10-4.

Answer 1

Answer: pH of solution is = 1.82

Explanation:

$HNO_2\rightarrow H^+NO_2^-$

 cM              0             0

$c-c\alpha$        $c\alpha$          $c\alpha$  

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.500 M and $\alpha$ = ?

$K_a=4.6\times 10^{-4}$

Putting in the values we get:

$4.6\times 10^{-4}=\frac{(0.500\times \alpha)^2}{(0.500-0.500\times \alpha)}$

$(\alpha)=0.030$

$[H^+]=c\times \alpha$

$[H^+]=0.500\times 0.030=0.015$

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

$pH=-log[0.015]=1.82$

Thus pH of a 0.500 M $HNO_2$ solution is 1.82