Answer:
Step-by-Step Explanation:
The principal rules include:
a) Bases contains higher pH compared to acids
b) The stronger acid contains reduced pH.
c) Stronger acid contains higher Ka. Hence, acids possessing higher Ka contain lower pH.
d) Stronger base contains higher Kb. Hence, bases possessing higher Kb contain higher pH.
Combining all these rules and putting them into action; in order of increasing pH, we have:
HF (1); HNO2 (2); NH3 (3); CH3NH2 (4)
For the second part:
a) NaCl occurs to be the salt of both strong acid & strong base. For that reason, it is a neutral salt with a pH of 7.0
b) NaF is a basic salt because, it serves as salt of strong base as seen in NaOH and that of a weak acid(e.g HF).
However, it produces Na^+ & F^- ions when it exist in solution.
where;
F^- = strong conjugate base for a weak acid HF.
Thus, the Kb of [tex]F^-[/tex] = [tex]\dfrac{10^{-14} }{6.8 \times 10^{-4}}[/tex] = [tex]1.47 \times 10^{-11}[/tex]
Now, the [tex][OH^-][/tex] in 0.1 M of NaF solution = [tex](0.1 \times 1.47 \times 10^{-11} ) 0.5[/tex]
[tex]= 1.21 \times 10^{-6} \ M[/tex]
[tex]pOH = -log [OH^-][/tex]
[tex]pOH = -log (1.21 \times 10^{-6} )[/tex]
[tex]pOH = 5.92 \\ \\ pH = 14 - pOH \\ \\ pH = 14 - 5.92 \\ \\ \mathbf{pH = 8.08}[/tex]
c) [tex]KNO_2[/tex] occurs to be a basic salt because it is a salt of both strong base (KOH) and a weak acid [tex](HNO_2).[/tex]
yields in solution.
Kb of [tex]NO^{2-}[/tex] =[tex]\dfrac{10^{-14}}{(4.5 \times 10^{-4})}[/tex] = [tex]2.2 \times 10^{-11}[/tex]
Now, the [tex][OH^-][/tex] in 0.1 M of [tex]KNO_2[/tex] solution = [tex](0.1 \times 2.22 \times 10^{-11})\times 0.5[/tex]
[tex]= 1.49 \times 10^{-6} \ M[/tex]
[tex]pOH = -log[OH^-][/tex]
[tex]pOH = -log (1.49 \times 10^{-6} )[/tex]
[tex]pOH = 5.83\\ \\ pH = 14 - pOH \\ \\ pH = 14 - 5.83 \\ \\ \mathbf{pH = 8.17}[/tex]
d)
NH_4Cl occurs to be an acidic salt because it is a salt of both weak base and a strong acid (HCl)
NH_4Cl yields [tex]NH_4^+[/tex] and [tex]Cl^-[/tex] when present in a solution.
Ka of [tex]NH_4^+[/tex] = [tex]\dfrac{10^{-14}}{1.8 \times 10^{-5}}[/tex]
[tex]= 5.56 \times 10^{-10}[/tex]
Now, the [tex][H^+] = (0.1 \times 5.56 \times 10^{-10}) 0.5[/tex]
[tex]= 7.45 \times 10^{-6}[/tex]
[tex]\mathbf{= 5.12}[/tex]
Thus, the order of the pH is:
NH_4Cl (1); NaCl (2); NaF (3); KNO_2 (4)
