Question

Calculate \(\left[\mathrm{H}^{+}\right]\) and the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of chloroacetic acid \(\left(K_{\mathrm{a}}=1.5 \times 10^{-3}\right)\).

Short answer

The concentration of hydrogen ions in the 0.10 M solution of chloroacetic acid is approximately 3.9 x 10^(-3) M. The pH of the solution is approximately 2.41.

Step-by-step solution

Write the ionization equation for chloroacetic acid

Chloroacetic acid (CICH2COOH) is a weak acid that partially ionizes in water. The ionization reaction can be represented as follows: $$\mathrm{ClCH_2COOH}\leftrightharpoons\mathrm{ClCH_2COO}^{-}+\mathrm{H}^{+}$$

Write the Ka expression for chloroacetic acid

The equilibrium constant for this weak acid ionization reaction, known as Ka, is given by: $$K_a = \frac{[\mathrm{ClCH_2COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{ClCH_2COOH}]}$$

Create an ICE table to relate the initial and equilibrium concentrations

An ICE table (initial, change, equilibrium) can help us relate the initial concentrations of the species involved in the ionization reaction to their equilibrium concentrations. Let x be the amount of \(\mathrm{H}^{+}\) ions produced, then: | | ClCH2COOH | ClCH2COO- | H+ | |-----------|-----------|-----------|-----| | Initial | 0.10 M | 0 | 0 | | Change | -x | +x | +x | | Equilibrium | 0.10 - x | x | x |

Rewrite the Ka expression using the ICE table values and solve for x

Substitute the equilibrium concentrations from the ICE table into the Ka expression: $$1.5\times10^{-3} = \frac{x^{2}}{0.10 - x}$$ To simplify this equation, we can make the assumption that x will be small compared to 0.10 (since the acid is weak). In that case: $$1.5\times10^{-3} \approx \frac{x^{2}}{0.10}$$ Now, we can solve for x: $$x^{2} \approx 1.5\times10^{-4}$$ $$x \approx \sqrt{1.5\times10^{-4}}$$ $$x \approx 3.9\times10^{-3}$$

Calculate the hydrogen ion concentration

Since x represents the equilibrium concentration of hydrogen ions, we have: $$[\mathrm{H}^{+}] = 3.9\times10^{-3}\mathrm{M}$$

Calculate the pH of the solution

The pH is defined by the negative logarithm of the hydrogen ion concentration: $$\mathrm{pH} = -\log([\mathrm{H}^{+}])$$ $$\mathrm{pH} = -\log(3.9\times10^{-3})$$ $$\mathrm{pH} \approx 2.41$$ Thus, the concentration of hydrogen ions \([\mathrm{H}^{+}]\) in the \(0.10\mathrm{M}\) solution of chloroacetic acid is \(3.9\times10^{-3}\mathrm{M}\), and the pH of the solution is approximately 2.41.