Question

A buffer is prepared by dissolving \(0.062 \mathrm{~mol}\) of sodium fluoride in \(127 \mathrm{~mL}\) of \(0.0399 \mathrm{M}\) hydrofluoric acid. Assume no volume change after \(\mathrm{NaF}\) is added. Calculate the \(\mathrm{pH}\) of this buffer.

Short answer

Answer: The pH of the buffer solution is 5.80.

Step-by-step solution

Calculate the molar concentration of HF (weak acid)

Given the volume of the solution, \(127 \mathrm{~mL}\), and the molarity of HF, \(0.0399 \mathrm{M}\), we can calculate the moles of HF: Moles of HF = Molarity × Volume = \(0.0399 \mathrm{M} × 127 \mathrm{~mL} × \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.00507 \mathrm{~mol}\) Now, we can find the molar concentration of HF in the final buffer solution: \(\mathrm{[HF]} = \frac{\text{moles of HF}}{\text{total volume in L}} = \frac{0.00507 \mathrm{~mol}}{127 \times 10^{-3}\mathrm{L}} = 0.0399 \mathrm{M}\)

Calculate the molar concentration of F- (conjugate base)

It is given that there are \(0.062 \mathrm{~mol}\) of sodium fluoride (NaF) in the buffer solution. Sodium fluoride (NaF) dissociates completely in water to form \(\mathrm{F^-}\): \(\mathrm{NaF} \rightarrow \mathrm{Na^+} + \mathrm{F^-}\) Therefore, the moles of \(\mathrm{F^-}\) in the buffer solution are equal to the moles of \(\mathrm{NaF}\): Moles of F- = \(0.062 \mathrm{~mol}\) Now, we can find the molar concentration of F- in the final buffer solution: \(\mathrm{[F^-]} = \frac{\text{moles of F^-}}{\text{total volume in L}} = \frac{0.062\mathrm{~mol}}{127 \times 10^{-3}\mathrm{L}} = 0.488 \mathrm{M}\)

Determine the Ka (acid dissociation constant) of HF

To calculate the pH of this buffer, we need the acid dissociation constant, Ka, for hydrofluoric acid (HF): \(\mathrm{HF} \leftrightarrows \mathrm{H^+} + \mathrm{F^-}\) We can find the Ka of HF from any reliable resource (e.g., Chemistry textbook, online sources). The Ka of HF is approximately \(6.6 \times 10^{-4}\).

Apply the Henderson-Hasselbalch equation to find pH

We can now use the Henderson-Hasselbalch equation, which relates the pH, pKa(observed with the logarithm of Ka), and concentrations of the weak acid and its conjugate base: \(\mathrm{pH} = \mathrm{pKa} + \log \frac{\mathrm{[F^-]}}{\mathrm{[HF]}}\) First, we need to find the pKa value: \(\mathrm{pKa} = -\log(\mathrm{Ka}) = -\log(6.6 \times 10^{-4}) \approx 3.18\) Now, we can substitute the values into the Henderson-Hasselbalch equation: \(\mathrm{pH} = 3.18 + \log(\frac{0.488}{0.0399}) \approx 3.18 + 2.62 = 5.80\) Therefore, the pH of the buffer solution is 5.80.