A 0.100\(M\) solution of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) is 11.0\(\%\) ionized. Using this information, calculate \(\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right],\) and \(K_{a}\) for chloroacetic acid.

From the given information, we know that 11% of chloroacetic acid ionizes. We can use this information to calculate the equilibrium concentrations of ClCH₂COO⁻ and H⁺. If the 0.1M solution is 11.0% ionized, then: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right]_{eq} = \left[\mathrm{H}^{+}\right]_{eq} = 0.11 \times 0.100 M = 0.0110 M\]

Next, let's determine the equilibrium concentration of ClCH₂COOH. Since 11% of it ionized, then 100% - 11% = 89% of it remains un-ionized in the equilibrium: \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right]_{eq} = 0.89 \times 0.100 M = 0.0890 M\]

Now that we have the equilibrium concentrations, we can use them to find the Ka of chloroacetic acid. The ionization dissociation reaction of chloroacetic acid can be written as: \[\mathrm{ClCH}_{2}\mathrm{COOH} \rightleftharpoons \mathrm{ClCH}_{2}\mathrm{COO}^{-}+\mathrm{H}^{+}\] The Ka expression for this reaction is: \[K_{a}=\frac{[\mathrm{ClCH}_{2}\mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{ClCH}_{2}\mathrm{COOH}]} \] Inserting the calculated equilibrium concentrations: \[K_{a}=\frac{(0.0110)(0.0110)}{(0.0890)} = 1.36 \times 10^{-3}\] We have obtained the equilibrium concentrations for ClCH₂COO⁻, H⁺ and ClCH₂COOH, as well as the Ka value. The results are as follows: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right]_{eq}=0.0110 M\] \[\left[\mathrm{H}^{+}\right]_{eq}=0.0110 M\] \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right]_{eq}=0.0890 M\] \[K_{a}=1.36 \times 10^{-3}\]