Calculate the pH of a solution formed by mixing 100.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M}\) NaF and 100.0 \(\mathrm{mL}\) of 0.025 \(\mathrm{M} \mathrm{HCl} .\)

First, we need to calculate the moles of NaF and HCl in the original solutions. To do this, use the formula: moles = molarity × volume For NaF: moles NaF = \(0.100 \, \mathrm{M}\) × \(100.0 \, \mathrm{mL}\) moles NaF = \(0.100 \, \mathrm{M}\) × \(0.100 \, \mathrm{L}\) moles NaF = 0.0100 moles For HCl: moles HCl = \(0.025 \, \mathrm{M}\) × \(100.0 \, \mathrm{mL}\) moles HCl = \(0.025 \, \mathrm{M}\) × \(0.100 \, \mathrm{L}\) moles HCl = 0.00250 moles

When mixing the solutions, NaF, a salt, will dissociate into Na+ and F- ions, while HCl, a strong acid, will dissociate into H+ and Cl- ions. The H+ ions from HCl and the F- ions from NaF will then react to form HF, according to the following equation: H+ + F- -> HF Since there are more F- ions than H+ ions, after the reaction, we will have an excess of F- ions left. In order to find the remaining F- ions, subtract the moles of H+ ions from the initial moles of F- ions. Remaining F- ions: moles F- = moles NaF - moles HCl moles F- = 0.0100 - 0.00250 = 0.00750 moles

Now, we need to calculate the concentration of the F- ions and [HF] after mixing the solutions. Since the volume of the mixed solutions is 200.0 mL (100.0 mL from NaF and 100.0 mL from HCl), we can calculate the concentrations using the following formula: concentration = (moles) / (final volume in liters) For F- ions: concentration F- = 0.00750 / 0.200 = \(0.0375 \, \mathrm{M}\) For HF: Since the moles of HF formed are equal to the moles of HCl consumed (0.00250 moles), the concentration of HF is: concentration HF = 0.00250 / 0.200 = \(0.0125 \, \mathrm{M}\)

Now we can use the Ka expression for HF (Ka = [H+][F-] / [HF]) to calculate the concentration of H+ ions and determine the pH of the solution. The Ka for HF is 6.6 x 10^(-4). \(6.6 \times 10^{-4} = \frac{[H^+][0.0375]}{0.0125 - [H^+]}\) We can solve for [H+] using the quadratic formula or make an approximation (assuming that [H+] is much smaller than 0.0125) and obtain: >H+ concentration = \(6.6 \times 10^{-4} \cdot \frac{0.0125}{0.0375}\) >[H+] = \(1.65 \times 10^{-4} \mathrm{M}\) Now, we can use the pH formula to calculate the pH of the solution: pH = -log10[H+] pH = -log10(\(1.65 \times 10^{-4}\)) pH ≈ 3.78 The pH of the solution formed by mixing 100.0 mL of 0.100 M NaF and 100.0 mL of 0.025 M HCl is approximately 3.78.