Question

Consider a solution initially containing 100. mmol fluoride anion and 100. mmol of hydrogen fluoride (HF) in 1.0 L of solution. What is the pH after addition of 15.9 mmol of HCl to this solution? Assume that the volume change upon addition of acid is negligible.

HF, Ka = 3.50 × 10−4

Answer 1

F^-   + HCl    HF + Cl^-

15.9 mmol of HCl will react with 15.9 mmol of fluoride anion to give 15.9 mmol of HF.

Thus, the new mol of HF after the addition of HCl = (100. + 15.9) = 116. mmol = 116. x 10^-3 mol = 0.116 mol

And the new mol of fluoride anion after the addition of HCl = (100. - 15.9) = 84.1 mmol = 0.084 mol

Volume of the solution = 1.0 L

The the new concentration of HF after the addition of HCl = 0.116/1.0 = 0.116 M

And the new concentration of fluoride anion after the addition of HCl = 0.084/1.0 = 0.084 M

Given, K_a of HF = 3.50 × 10⁻⁴

Now, from Henderson-Hasselbalch equation

pH = pK_a + log [base]/[acid]

or, pH = - log K_a + log [F-]/[HF]

          = - log(3.50 × 10⁻⁴) + log (0.084)/(0.116)

          = 3.46 - 0.14

          = 3.32

Hence, the pH after addition of 15.9 mmol of HCl to this solution = 3.32