Sodium fluoride is added to many municipal water supplies to reduce tooth decay.   Calculate the pH of a 3.39×10-3 M  solution of NaF, given that the Ka of HF = 6.80 x 10-4 at 25°C.

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Sodium fluoride is added to many municipal water supplies to reduce tooth decay.
 
Calculate the pH of a 3.39×10-3 M  solution of NaF, given that the Ka of HF = 6.80 x 10-4 at 25°C.
Expert Solution
Given

The dissociation of NaF is given as-

NaF Na++F

The concentration of NaF, [NaF]=3.39 ×10-3 M[NaF]=0.00339 M

Ka of HF = 6.80 x 10-4

Hydrolysis

As the reaction happens in an aqueous medium the water molecules dissociate in the form-

H2O H+ + OH-

Na+ only reacts with OH- ions in the reaction mixture which is a strong base and dissociates completely and will not hydrolyze, whereas on the other hand, F- can only react with H+ ions and form HF which is a weak acid and will hydrolyze. The equation for the hydrolysis is given as-

F+ H2O  HF + OH

As the conc. of F- is the same as that of NaF given because NaF dissociates completely in the aqueous solution. So, the hydrolysis equation is an equilibrium equation which means not all F- gives HF but the pH is given by the concentration of OH- ions in the solution which is given as-

[OH]=Kb.[A]where, [A] = conc. of F- ions

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