A \(0.100 \mathrm{M}\) solution of bromoacetic acid $\left(\mathrm{BrCH}_{2} \mathrm{COOH}\right)$ is 13.2\% ionized. Calculate $\left[\mathrm{H}^{+}\right],\left[\mathrm{BrCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{BrCH}_{2} \mathrm{COOH}\right]$ and \(K_{a}\) for bromoacetic acid.

The initial concentration is given as \(0.100\,\mathrm{M}\).

We are given that the solution is 13.2% ionized. We can calculate the concentration of dissociated bromoacetic acid by multiplying the initial concentration by the percent ionization: \(\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right] = 0.100\,\mathrm{M} \times 0.132 = 0.0132\,\mathrm{M}\)

Since the solution is 13.2% ionized, we can say that it still contains 86.8% of the initial bromoacetic acid. We can calculate the concentration of the undissociated bromoacetic acid by multiplying the initial concentration by the remaining percent: \(\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right] = 0.100\,\mathrm{M} \times 0.868 = 0.0868\,\mathrm{M}\)

Since the concentration of hydrogen ions is equal to the concentration of bromoacetate ions, the concentration of hydrogen ions is: \(\left[\mathrm{H}^{+}\right] = 0.0132\,\mathrm{M}\)

Using the ionization equation of a weak acid, we can calculate the \(K_{a}\) as follows: \(K_{a} = \frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right]}{\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right]}\) Substitute the values obtained in Steps 2, 3 and 4 into the above equation: \(K_{a} = \frac{0.0132\,\mathrm{M}\times 0.0132\,\mathrm{M}}{0.0868\,\mathrm{M}}= 2.0 \times 10^{-3}\) Thus, the concentrations of the ions and the undissociated bromoacetic acid are: \(\left[\mathrm{H}^{+}\right] = 0.0132\,\mathrm{M}\) \(\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right] = 0.0132\,\mathrm{M}\) \(\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right] = 0.0868\,\mathrm{M}\) and the \(K_{a}\) for bromoacetic acid is: \(K_{a} = 2.0 \times 10^{-3}\)