Determine the ph of a 0.22 m naf solution at 25°
c. the salt completely dissociates into na (aq) and f- (aq), and the na (aq) ion has not acid or base properties. the ka of hf is 3.5 × 10-5.

Doing the ICE approache
                    naf   --->  na    +   f-
Initial           0.22 
Change      -0.22         0.22     0.22
Equilibrium  0              0.22      0.22

Hydrolysis will follow the dissociation
                      f-  + H2O >   hf  + OH-
Initial            0.22
Change        -x                          x        x
Equilibrium    0.22-x                  x       x

We are given with the ka of hf
hf = 3.5x10^-5 = x(x)/(0.22-x)
Solving for x
x = 2.76x10^-3 m which is the concentration of OH- ions

Solving for pH
pH = 14 - pOH
pH = 14 - (-log [2.76x10^-3])
pH = 11.44
The pH is 11.44 which means that the solution is basic.

Woah that’s a lot to read xd i’ll try though. i’ll get back t you if. i know the andwer