Question

Question 5 A solution is prepared at 25°C that is initially 0.35M in chlorous acid HClO2, a weak acid with =Ka×1.110−2, and 0.29M in sodium chlorite NaClO2. Calculate the pH of the solution. Round your answer to 2 decimal places.

Answer 1

Answer : The pH of the solution is, 1.88

Explanation : Given,

$K_a=1.1\times 10^{-2}$

Concentration of $HClO_2$ = 0.35 M

Concentration of $NaClO_2$ = 0.29 M

First we have to calculate the value of $pK_a$.

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log [K_a]$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (1.1\times 1-^{-2})$

$pK_a=2-\log (1.1)$

$pK_a=1.96$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[NaClO_2]}{[HClO_2]}$

Now put all the given values in this expression, we get:

$pH=1.96+\log (\frac{0.29}{0.35})$

$pH=1.88$

Therefore, the pH of the solution is, 1.88