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? ( ans: 2.0 × 10^-9)

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.

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it changes food into energy that the organism can use.

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the cell will not be replaced when it is damaged.

explanation: