A $100.0 \mathrm{~mL}$ sample of $0.20 \mathrm{M} \mathrm{HF}$ is titrated with $0.10 \mathrm{M} \mathrm{KOH}$. Determine the $\mathrm{pH}$ of the solution after the addition of 200.0 $\mathrm{mL}$ of $K O H$. The $K_{\mathrm{a}}$ of HF is $3.5 \times 10^{-4}$. 8.14 7.00 10.54 3.46 9.62

Step 1 ：Firstly, we find the moles of HF and KOH. The moles of HF is given by the product of volume and molarity, which is \(0.20 \times 100.0 = 20.0\) millimoles. Similarly, the moles of KOH is \(0.10 \times 200.0 = 20.0\) millimoles.

Step 2 ：Since the moles of HF and KOH are equal, all the HF will react with KOH to form F- and H2O. Therefore, the solution is now a solution of F- ions.

Step 3 ：The F- ions will react with water to form HF and OH-. This reaction is governed by the Kb of F-, which is given by \(K_w / K_a\), where \(K_w = 1.0 \times 10^{-14}\) and \(K_a = 3.5 \times 10^{-4}\). Therefore, \(K_b = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}} = 2.86 \times 10^{-11}\).

Step 4 ：Since the reaction of F- with water is a weak base reaction, we can use the expression for Kb to find the concentration of OH-. Let x be the concentration of OH-. Then, \(K_b = [OH-][HF]/[F-] = x^2 / (20.0 - x)\). Solving this equation for x, we find that x is approximately equal to \(\sqrt{20.0 \times 2.86 \times 10^{-11}} = 2.68 \times 10^{-6}\) M.

Step 5 ：Finally, we can find the pH of the solution by first finding the pOH, which is given by \(-\log[OH-] = -\log(2.68 \times 10^{-6}) = 5.57\). Then, the pH is given by \(14 - pOH = 14 - 5.57 = \boxed{8.43}\).