(a) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid in a solution containing $0.050 \mathrm{M}$ sodium lactate.

We can represent the ionization of lactic acid (HA) as follows: \(HA \rightleftharpoons H^+ + A^-\) Given the initial concentration of lactic acid is 0.250 M and Ka = \(1.4 \times 10^{-4}\). Construct an ICE table for the equilibrium concentrations: Initial 0.250 M 0 0 Change -x +x +x Equilibrium 0.250-x x x Now, use the Ka expression to solve for x: \[K_a =\frac {[H^+][A^-]}{[HA]} = \frac {x \times x}{0.250-x}\] Solve the equation for x: \(1.4 \times 10^{-4} = \frac {x^2}{0.250-x}\) \(x\) is the concentration of H⁺, as the solution mostly dissociates to a limited extent, we can assume that \(x << 0.250\) Simplify and solve for x: \(1.4 \times 10^{-4} ≈ \frac {x^2}{0.250}\) \(x ≈ \sqrt{0.250 \times 1.4 \times 10^{-4}}\) \(x ≈ 0.00295\,\mathrm{M}\) Calculate the percent ionization: \(\% \textrm{ ionization } = \frac{[H^+]}{0.250\,\mathrm{M}} \times 100\) \(\% \textrm{ ionization } ≈ \frac{0.00295\,\mathrm{M}}{0.250\,\mathrm{M}} \times 100\) \(\% \textrm{ ionization } ≈ 1.18\%\)

Now we have a buffer solution containing both the weak acid and its salt. In this case, the initial condition changes: Initial 0.250 M 0.050 M 0.050 M Change -y y y Equilibrium 0.250-y 0.050+y 0.050+y Substitute the equilibrium concentrations into the Ka expression: \(1.4 \times 10^{-4} = \frac{((0.050+y)(y))}{(0.250-y)}\) Assuming the change is small, neglect y compared to 0.050 and 0.250: \(1.4 \times 10^{-4} ≈ \frac{0.050 \times y}{0.250}\) Solve for y: \(y ≈ \frac{1.4 \times 10^{-4} \times 0.250}{0.050}\) \(y ≈ 0.0007\,\mathrm{M}\) Calculate the percent ionization: \(\%\, \textrm{ionization} = \frac{[H^+]}{0.250\,\mathrm{M}} \times 100\) \(\%\, \textrm{ionization} ≈ \frac{0.0007\,\mathrm{M}}{0.250\,\mathrm{M}} \times 100\) \(\%\, \textrm{ionization} ≈ 0.28\%\)