Question

The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0°c is 4.48. what is the value of ka for hbro?

Answer 1

The solution would be like this for this specific problem:

Given:

 

pH of a 0.55 M hypobromous acid (HBrO) at 25.0 °C =  4.48

 

[H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-] 

Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9

 

To add, Hypobromous Acid does not require acid adjustment, which is necessary for chlorine-based product and is stable and effective in pH ranges of 5-9.

Hypobromous Acid combines with organic compounds to form a bromamine. Chlorine also combines with the same organic compounds to form a chloramine. It is also one of the least expensive intervention antimicrobial compounds available.

Answer 2

Answer:

The value of dissociation constant for the hypobromous acid is $1.993\times 10^{-9}$.

Explanation:

The pH of the solution = 4.48

Concentration of hypobromous acid,$[HBrO]=c=0.55 M$

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

                           $HBrO\rightleftharpoons OBr^-+H^+$

initially conc.         c                       0         0

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

First we have to calculate the concentration of value of degree of dissociation $\alpha$.

Expression for dissociation constant is given as:

$k_a=\frac{(c\alpha )(c\alpha )}{c(1-\alpha )}=\frac{c(\alpha )^2}{(1-\alpha )}$..(1)

$[H^+]=c\alpha$

$pH=4.48=-\log[H^+]=-\log[c\alpha ]=-\log[0.55 M\times \alpha ]$

$\alpha =6.0205\times 10^{-5}$

Substituting all the values in (1), we get the value of dissociation constant:

$K_a=\frac{0.55 M(6.0205\times 10^{-5})^2}{(1-(6.0205\times 10^{-5}))}$

$K_a=1.993\times 10^{-9}$

The value of dissociation constant for the hypobromous acid is $1.993\times 10^{-9}$.