A solution is made by adding 50.0 \(\mathrm{mL}\) of 0.200\(M\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) to 50.0 \(\mathrm{mL}\) of \(1.00 \times 10^{-3} \mathrm{M} \mathrm{HCl}\) a. Calculate the pH of the solution. b. Calculate the acetate ion concentration.

To calculate the moles of each acid, we will use the formula: moles = volume (in L) × molarity. The volume of both solutions is 50.0 mL, so we need to convert this to liters. Volume in L = volume in mL/1000 = 50.0 / 1000 = 0.050 L Moles of acetic acid = volume in L × molarity = 0.050 L × 0.200 M = 0.010 moles Moles of hydrochloric acid (HCl) = volume in L × molarity = 0.050 L × 0.00100 M = 0.000050 moles

Since HCl is a strong acid, it will completely ionize, and all of its protons will react with the acetic acid in the solution. The moles of HCl will decrease to zero, and the moles of acetic acid will be reduced by an equal amount. Moles of acetic acid after reaction = 0.010 moles - 0.000050 moles = 0.00995 moles Moles of HCl after reaction = 0.000050 moles - 0.000050 moles = 0 moles

The ionization of acetic acid can be represented by the following equilibrium equation: CH₃COOH + H₂O ↔ CH₃COO⁻ + H₃O⁺ Ka = \([CH₃COO⁻][H₃O⁺]\) / \([CH₃COOH]\) Given Ka = \(1.8 \times 10^{-5}\), we can set up an ICE (Initial, Change, Equilibrium) table to find the hydrogen ion concentration in the solution. [CH₃COOH] [CH₃COO⁻] [H₃O⁺] Initial 0.00995 0 0 Change -x x x Equil. 0.00995-x x x Using the Ka expression, \(1.8 \times 10^{-5}\) = \(\frac{x^2}{0.00995-x}\) Since Ka is very small, we can assume x is negligible compared to 0.00995. Thus, the equation can be simplified to: \begin{align*} 1.8 \times 10^{-5} &= \frac{x^{2}}{0.00995} \\ x^{2} &= 1.8 \times 10^{-5} \times 0.00995 \\ x &= \sqrt{1.8 \times 10^{-5} \times 0.00995} \\ x &= \text{1.3438} \times 10^{-4} \end{align*} The equilibrium concentration of hydrogen ions, [H₃O⁺], is \(1.3438 \times 10^{-4} M\).

Now that we have the hydrogen ion concentration, we can calculate the pH using the formula: pH = -log [H₃O⁺]. pH = -log( \(1.3438 \times 10^{-4}\)) = 3.87 The pH of the solution is approximately 3.87.

We determined the equilibrium concentration of CH₃COO⁻ (x) in Step 3. The acetate ion concentration is equal to x: [CH₃COO⁻] = \(1.3438 \times 10^{-4} M\) The acetate ion concentration in the solution is approximately \(1.3438 \times 10^{-4} M\).