To calculate equivalence point pH, you have to realize that two things have happened:
HF + OH- | ![]() | H2O + 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.
F- + H2O | ![]() | OH- + HF |
molarity of HF |
molarity of OH- |
molarity 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-]).
Copyright © 1997-2010 by Fred Senese
Comments & questions to fsenese@frostburg.edu
Last Revised 02/23/18.URL: http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/print-equivalence-point-HF-NaOH.shtml