Question

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.

Answer 1

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.