Calculate the \(\mathrm{pH}\) of a solution prepared by mixing \(250 . \mathrm{mL}\) of \(0.174 m\) aqueous \(\mathrm{HF}\) (density \(=1.10 \mathrm{~g} / \mathrm{mL}\) ) with \(38.7 \mathrm{~g}\) of an aqueous solution that is \(1.50 \% \mathrm{NaOH}\) by mass (density \(=\) \(1.02 \mathrm{~g} / \mathrm{mL}) \cdot\left(K_{\mathrm{a}}\right.\) for \(\left.\mathrm{HF}=7.2 \times 10^{-4} .\right)\)

First, we need to calculate the moles of the HF and NaOH present in the initial solutions. Moles of HF = Molarity × Volume = \(0.174 \mathrm{M} × 250 \mathrm{mL}\) (convert to L) = \(0.174 \mathrm{M} × 0.250 \mathrm{L} = 0.0435 \mathrm{mol}\) Next, we will find the moles of NaOH in the NaOH solution. Since the solution is 1.50% NaOH by mass, the mass of NaOH can be obtained as follows: Mass of NaOH = \((1.50 \%) × 38.7 \mathrm{g} = 0.58 \mathrm{~g}\) Now, we can calculate the moles of NaOH: Moles of NaOH = \(\frac{Mass}{Molar \thinspace Mass\thinspace of\thinspace NaOH} = \frac{0.58 \mathrm{g}}{40\thinspace \mathrm{g/mol}} = 0.0145\thinspace \mathrm{mol}\)

Now we need to determine which of the two reactants is the limiting reactant. In this case, it's the reactant present in the smallest relative amount: Limiting reactant = \(\min\Bigl(\frac{Moles\thinspace of\thinspace HF}{1},\frac{Moles\thinspace of\thinspace NaOH}{1}\Bigl) =\min(0.0435\thinspace \mathrm{mol}, 0.0145\thinspace \mathrm{mol}) = 0.0145\thinspace \mathrm{mol}\) of NaOH.

Once the limiting reactant is used up, calculate the moles of the remaining species: Moles of HF remaining = 0.0435 - 0.0145 = 0.0290 mol Moles of NaOH = 0 (used up completely) Moles of H2O = 0.0145 mol (formed during the reaction)

Now, we need to determine the new concentrations of HF and its conjugate base (F-) in the resulting solution and calculate the new volume of the solution. The volume of HF solution: \(V_{HF} = 250 \mathrm{mL} = 0.250 \mathrm{L}\) The volume of NaOH solution: \(V_{NaOH} = \frac{38.7 \mathrm{g}}{1.02 \mathrm{g/mL}} = 37.94 \mathrm{mL} = 0.0379 \mathrm{L}\) Total volume: \(V = 0.250 + 0.0379 = 0.2879 \mathrm{L}\) The new concentration of HF: \([\mathrm{HF}] = \frac{0.0290\thinspace \mathrm{mol}}{0.2879\thinspace \mathrm{L}} = 0.1007\thinspace \mathrm{M}\) The new concentration of F-: \([\mathrm{F}^-] = \frac{0.0145\thinspace \mathrm{mol}}{0.2879\thinspace \mathrm{L}} = 0.0503\thinspace \mathrm{M}\)

To calculate the pH of the resulting solution, we need to consider the ionization of HF, which is a weak acid: HF + H2O <-> H3O+ + F- As HF is a weak acid its ionization constant, \(K_a\), will be given by: \(K_a = \frac{[\mathrm{H_3O^+}][\mathrm{F}^-]}{[\mathrm{HF}]}\) Now, we can use the concentrations of HF and F- calculated in Step 4 and the given \(K_a\) value (\(7.2 \times 10^{-4}\)) to find the pH. Let \(x\) be the concentration of \(\mathrm{H_3O^+}\). Then, the ionization can be expressed as: \(K_a = \frac{x \cdot [\mathrm{F}^- + x]}{[\mathrm{HF}] - x} = \frac{x \cdot (0.0503 + x)}{(0.1007 - x)}\) As HF is a weak acid, \(x\) will be much smaller than \(0.1007\), so we can approximate \(x \ll 0.1007\). Then the equation will be: \(K_a = \frac{x \cdot 0.0503}{0.1007 - x} \approx \frac{x \cdot 0.0503}{0.1007}\) Now, we can solve for x: \(x = [\mathrm{H_3O^+}] = \frac{K_a \cdot [\mathrm{HF}]}{[\mathrm{F}^-]} = \frac{(7.2 \times 10^{-4})(0.1007)}{0.0503} = 0.00142\) Finally, we can calculate the pH of the solution: \(\mathrm{pH} = -\log([\mathrm{H_3O^+}]) = -\log(0.00142) \approx 2.85\) So the pH of the resulting solution is approximately 2.85.