Respuesta :
Answer:
You are already giving the answer, see procedure and work below
Explanation:
To know this, let's do it by parts. First, let's calculate the concentration of H+ that belongs to the pH of 4.48.
We know that pH = -log[H+], so if we solve for [H+]:
[H+] = antlog(-pH)
Solving for [H+]:
[H+] = antlog(-4.48) = 3.31x10^-5 (I used a calculator to know this value)
Now that we have the concentration of H+ in equilibrium, let's do the equation for this reaction:
HBrO <---------> H+ + BrO-
This is the general equation. At the beggining we only have 0.55 M of HBrO, and in equilibrium we have the following. (I'll put an i, for the innitial reaction and a e, for equilibrium)
HBrO <---------> H+ + BrO-
i) 0.55 0 0
e) 0.55 - x x x
The value of "x" here, would be the concentration value in equilibrium, and this value belongs to the pH of the solution. This value is 3.31x10^-5. So, to get the value of Ka we need to write the expression for equilibrium:
Ka = [products] / [Reactants]
In this case, products would be H+ and BrO+, and reactants only HBrO so:
Ka = x^2 / 0.55 - x
However, we already know the value of x, so replacing here we have:
Ka = (3.31x10^-5)^2 / 0.55 - 3.31x10^-5
Now we have to solve for Ka:
Ka = 1.096x10^-9 / 0.5499
Ka = 1.99x10^-9 which you can round to 2x10^-9.
