The pH of a \(0.063-M\) solution of hypobromous acid (HOBr but usually written \(\mathrm{HBrO}\) ) is \(4.95 .\) Calculate \(K_{\mathrm{a}} .\)

To find the concentration of H+ ions, we can use the formula: \[ \mathrm{pH} = -\log_{10} [\mathrm{H}^+]\] We can rearrange the formula to find the concentration of H+ ions: \[ [\mathrm{H}^+] = 10^{-\mathrm{pH}}\] Now, we have the pH value (4.95). Let's plug that in the formula to find the concentration of hydrogen ions: \[ [\mathrm{H}^+] = 10^{-4.95}\]

Since the hypobromous acid (HBrO) dissociates as follows: \[ \mathrm{HBrO} \rightleftharpoons \mathrm{H}^+ + \mathrm{BrO}^-\] We can assume that the initial concentration of BrO- ions is the same as the concentration of hydrogen ions formed during dissociation since one molecule of HBrO generates one molecule of BrO-. Therefore, the concentration of BrO- ions is equal to the concentration of H+ ions. \[ [\mathrm{BrO}^-] = [\mathrm{H}^+]\] Using the value of [H+] we found in step 1, we can determine the concentration of BrO- ions.

Now, we have all the necessary values for calculating the Ka of hypobromous acid. The equilibrium expression for the dissociation of HBrO in water is: \[K_a = \frac{[\mathrm{H}^+] [\mathrm{BrO}^-]}{[\mathrm{HBrO}]}\] Note that we were given the initial concentration of HBrO (0.063 M). As the dissociation takes place, it is assumed that the change in the concentration of HBrO is -x, while H+ and BrO- concentration increases by x. Therefore, the concentration of HBrO will be 0.063 - x. Using the equilibrium concentrations, plug in the values for H+ ions, BrO- ions, and the initial concentration of HBrO: \[K_{a}=\frac{([\mathrm{H}^+])([\mathrm{BrO^-}])}{0.063-x}\] Since x is very small compared to 0.063, we can consider (0.063 - x) to be approximately equal to 0.063: \[K_{a}=\frac{([\mathrm{H}^+])([\mathrm{BrO^-}])}{0.063}\] Now just insert the concentration values for H+ and BrO- ions calculated earlier and find the value of Ka for hypobromous acid.