Answer:
The value of dissociation constant for the hypobromous acid is [tex]1.993\times 10^{-9}[/tex].
Explanation:
The pH of the solution = 4.48
Concentration of hypobromous acid,[tex][HBrO]=c=0.55 M[/tex]
The equilibrium reaction for dissociation of HBrO (weak acid) is,
[tex]HBrO\rightleftharpoons OBr^-+H^+[/tex]
initially conc. c 0 0
At eqm. [tex]c(1-\alpha )[/tex] [tex]c\alpha [/tex] [tex]c\alpha [/tex]
First we have to calculate the concentration of value of degree of dissociation [tex]\alpha [/tex].
Expression for dissociation constant is given as:
[tex]k_a=\frac{(c\alpha )(c\alpha )}{c(1-\alpha )}=\frac{c(\alpha )^2}{(1-\alpha )}[/tex]..(1)
[tex][H^+]=c\alpha [/tex]
[tex]pH=4.48=-\log[H^+]=-\log[c\alpha ]=-\log[0.55 M\times \alpha ][/tex]
[tex]\alpha =6.0205\times 10^{-5}[/tex]
Substituting all the values in (1), we get the value of dissociation constant:
[tex]K_a=\frac{0.55 M(6.0205\times 10^{-5})^2}{(1-(6.0205\times 10^{-5}))}[/tex]
[tex]K_a=1.993\times 10^{-9}[/tex]
The value of dissociation constant for the hypobromous acid is [tex]1.993\times 10^{-9}[/tex].
A 0.55 M aqueous solution of HBrO has a pH of 4.48. The value of Ka for HBrO is 1.99 × 10⁻⁹.
HBrO is a weak acid according to the following equation.
HBrO ⇄ H⁺ + BrO⁻
Given the pH is 4.48, the concentration of H⁺ is:
[tex]pH = -log [H^{+} ]\\[H^{+} ] = antilog-pH = antilog-4.48 = 3.31 \times 10^{-5} M[/tex]
Given the initial concentration of the acid (Ca) is 0.55 M, we can calculate the acid dissociation constant (Ka) using the following expression.
[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(3.31 \times 10^{-5} )^{2} }{0.55} = 1.99 \times 10^{-9}[/tex]
A 0.55 M aqueous solution of HBrO has a pH of 4.48. The value of Ka for HBrO is 1.99 × 10⁻⁹.
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