Question

What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4

Answer 1

Answer:- pH = 8.08

Solution:- NaF is a salt of strong base and weak acid. So, the solution will be basic and the pH must be greater than 7. We will do the calculations to calculate the exact pH value for this. The ice table is made and then the calculations are done.

       $F^-(aq) + H_2O \leftrightharpoons HF(aq) + OH^-(aq)$

I              0.10                                                           0                  0

C              -X                                                           +X                +X

E          (0.10 - X)                                                      X                   X

Where X is the change in concentration.

The equilibrium expression for the equation is written as:

$Kb=\frac{[HF][OH^-]}{[F^-]}$

Kb could be calculated from the given Ka value as:

$Kb=\frac{1.0*10^-^1^4}{Ka}$

$Kb=\frac{1.0*10^-^1^4}{6.8*10^-^4}$

$Kb=1.5*10^-^1^1$

Let's plug in the values in the above equilibrium expression and solve it for X:

$1.5*10^-^1^1\frac{(x)^2}{0.10-x}$

Dissociation constant is very low and so the X on the bottom could be neglected and hence 0.10 - X could be taken as 0.10.

$1.5*10^-^1^1\frac{(x)^2}{0.10}$

On cross multiply:

$x^2=1.5*10^-^1^2$

Taking square root to both sides:

$x=1.22*10^-^6$

From ice table, hydroxide ion concentration is X.

So, $[OH^-]=1.22*10^-^6M$

$pOH=-log[OH^-]$

$pOH=-log1.22*10^-^6$

pOH = 5.91

pH = 14 - pOH

pH = 14 - 5.91

pH = 8.09

The closest choice is B, 8.08. So, the pH of the solution is 8.08.