Question

# Consider a solution initially containing 100. mmol fluoride anion and 100. mmol of hydrogen fluoride (HF)...

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

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, Ka of HF = 3.50 × 10−4

Now, from Henderson-Hasselbalch equation

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

or, pH = - log Ka + 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

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