Question

1) What is the pH of a 0.225 M aqueous solution of sodium fluoride, NaF? ____

The solution is:

A) Acidic

B) Basic

C) Neutral

2) What is the pH of a 5.86×10^-2 M aqueous solution of ammonium nitrate, NH₄NO₃ ? _______

The solution is:

A) Acidic

B) Basic

C) Neutral

Answer 1

1)

Solution :-

NaF gives the F^- which acts as base

F- + H2O   ------ > HF + OH-

lets calculate the concnetration of the OH- using the Kb

kb = Kw/ ka

Kb= 1*10^-14 / 6.8*10^-4

Kb = 1.47*10^-11

kb = [HF][OH-]/[F-]

1.47*10^-11 = [x][x]/[0.225]

1.47*10^-11* 0.225 = x^2

3.31*10^-12 = x^2

taking square root of the both sides we get

x= 1.82*10^-6 M

now lets calculate the pOh

pOH= -log [OH-]

pOH= -log [1.82*10^-6]

pOH = 5.74

now lets find the pH

pH+ pOH = 14

pH= 14 - pOH

pH= 14 - 5.74

pH= 8.26

So the pH of the solution is 8.49

It is basic