What is the pH at the equivalence point an HF/NaOH titration?

When 20 mL of a 1.5 moles/litre solution of HF is titrated with 1.0 moles/litre NaOH, what is the pH at the equivalence point?

I hope this is a hypothetical question. HF etches glass, and the fumes from a 1.5 M HF solution are deadly!

To calculate equivalence point pH, you have to realize that two things have happened:

  1. The acid is neutralized stoichimetrically. At equivalence all of the acid has reacted with the sodium hydroxide:
    HF + OH- H2O + F-
    At this point, there are no moles of HF left and (0.020 L)×(1.5 mol/L) = 0.030 mol of F-.

    Compute the molarity of the fluoride now. The volume of NaOH that was required to reach the equivalence point was (0.030 mol)/(1.0 mol NaOH/L) = 30 mL. The total volume of the solution at the equivalence point is 30 mL + 20 mL = 50 mL. So the molarity of F- is 0.030 mole/0.050 L or 0.6 M.

  2. The conjugate base hydrolyzes. The fluoride ion formed in step 1 reacts with water. This is called a basic hydrolysis. This isn't a complete reaction; you need to set up an equilibrium calculation here.
    F- + H2OOH- + HF
    The law of mass action for the hydrolysis is
    Kb = [HF][OH-]/[F-]
    You can get the equilibrium constant for the basic hydrolysis easily, if you have the acid dissociation constant: Kb=Kw/Ka. Organize the calculation as follows:
    of HF
    of OH-
    of F-
    Before Hydrolysis 0 ~0 0.6 M
    Change due to Hydrolysis +x +x -x
    After Hydrolysis +x +x 0.6 - x

    The values after hydrolysis are the equilibrium concentrations. Your next step is to use these values in the law of mass action to solve for x, the molarity of the hydroxide ion. The pH of the solution can easily be obtained from the hydroxide molarity since pH = -log[H+] = -log( Kw/[OH-]).

Author: Fred Senese senese@antoine.frostburg.edu

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