Question

calculate the pH of the following aqueous solution at 25 degreesCelsius and .35M NaF (pKa for HF=3.14)

Answer 1

Sol:-

NaF is a salt of weak acid HF and strong base NaOH , therefore resultant solution must be basic .

given pKa for HF=3.14

also we know that

pKa = -log Ka

Ka = 10^-pka

Ka = 10^-3.14

Ka = 7.2 x 10^-4

and

Kb = Kw / ka = 10^-14 / 7.2 x 10^-4 = 1.39 x 10^-11

Now ICE table for the given weak base F^- in aqueous solution is .

............... F^- (aq)..........+ ............H₂O (l) --------------> HF (aq) ............+ .............OH^- (aq)

I ............0.35 M ...........................................................0 M..................................0 M

C............. - x ............................................................... +x ................................... + x

E ........(0.35 -x )M.......................................................... x M ................................... x M

here x = degree of dissociation

Now the expression of Kb for this reaction is

Kb = [ HF] [OH^-] [ F^-]

1.39 x 10^-11 = x² / 0.35 - x

if x <<<0.35 , then neglect x , we have

x² = 0.35 x 1.39 x 10^-11

x = ( 4.86 x 10^-12 )^1/2

x = 2.2 x 10^-6

therefore

[ HF] = [OH^-] = x = 2.2 x 10^-6 M

and pOH = - log [OH^-]

pOH = - log 2.2 x 10^-6

pOH = - ( - 5.66 )

pOH = 5.66

Now pH + pOH = 14

so

pH = 14 - pOH

pH = 14 - 5.66

pH = 8.34

Hence pH is equal to 8.34