Question

Consider the titration of \(\mathrm{HF}\left(K_{\mathrm{a}}=6.7 \times 10^{-4}\right)\) with \(\mathrm{NaOH}\). What is the \(\mathrm{pH}\) when a third of the acid has been neutralized?

Short answer

Question: Calculate the pH when a third of an HF solution has been neutralized by NaOH, given the Ka value of HF is \(6.7 \times 10^{-4}\). Answer: The pH value depends on the initial concentration of the HF solution (\(C_{HF}\)). To calculate the pH when a third of HF has been neutralized, use the formula: pH = \(-\log_{10} \left(\sqrt{6.7 \times 10^{-4} \times \frac{2}{3}C_{HF}}\right)\).

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the titration of HF with NaOH is: HF + NaOH → NaF + H2O

Calculate the concentration of HF and NaOH

Let the initial concentration of HF be \(C_{HF}\). Since one-third of the acid has been neutralized by the base, two-thirds of the acid is still left. So, the new concentration of HF is \(\frac{2}{3}C_{HF}\). Since NaOH completely dissociates, its concentration at this stage becomes \(\frac{1}{3}C_{HF}\).

Write the dissociation equilibrium for HF

The dissociation of HF is represented as: HF <=> H+ + F- We are given the Ka value for HF: \(6.7 \times 10^{-4}\). Ka = \(\frac{[H^+] [F^-]}{[HF]}\) At equilibrium, let the concentration of H+ ions be x, so [F-] = x and [HF] = \(\frac{2}{3}C_{HF} - x\). So, the Ka equation becomes: \(6.7 \times 10^{-4} = \frac{x^2}{\frac{2}{3}C_{HF} - x}\)

Calculate the concentration of H+ ions and pH

Given that x is small compared to \(\frac{2}{3}C_{HF}\), the equation simplifies to: \(6.7 \times 10^{-4} = \frac{x^2}{\frac{2}{3}C_{HF}}\) Divide both sides by \(\frac{2}{3}C_{HF}\): \(x^2 = 6.7 \times 10^{-4} \times \frac{2}{3}C_{HF}\) \(x = \sqrt{6.7 \times 10^{-4} \times \frac{2}{3}C_{HF}}\) Now, we can find the pH using the H+ ion concentration: pH = \(-\log_{10} [H^+]\) pH = \(-\log_{10} \left(\sqrt{6.7 \times 10^{-4} \times \frac{2}{3}C_{HF}}\right)\) In this equation, the pH value depends on the initial concentration of HF (\(C_{HF}\)). Therefore, at any given initial concentration, we can now calculate the pH when a third of HF has been neutralized using the provided formula.