2SO_2(g) + O_2(g) $$\rightleftharpoons$$ 2SO_3(g),    K

SO_3(g) $$\rightleftharpoons$$ SO_2(g) + $${1 \over 2}$$ O_2(g),     K' = $$\sqrt {{1 \over K}} $$

$$ \therefore $$ K' = $$\sqrt {{1 \over {278}}} = $$ 6.0 $$ \times $$ 10^$$-$$2

For this example,

CH₃COOH ⇌ CH₃COO^– + H⁺ ;

CH₃COONa ⇌ CH₃COO^– + Na⁺

when few drops of HCl are added to this buffer, the H⁺ of HCl immediatly combine with CH₃COO^– ions to form undissociated acetic acid molecules. Thus there will be no H⁺ ions to combine with CH₃COO^– ions to form undissociated acetic acid molecules. Thus there will be no appreciable change in its pH value. Like wise if few drops of NaOH are added, the OH^– ions will combine with H⁺ ions to form unionised water molecule. Thus pH of solution will remain constant.

All of the given salts have same anion i.e., Cl which on hydrolysis gives HCl which is a strong acid.

Now, among the salts which have cation that gives a strongest base on hydrolysis of salt have the highest pH value. As Ba form Ba(OH)₂ which is a stronger base thus, it results in the highest pH value.

	Ba(OH)2	⇌	Ba2+	+	2OH
At equilibrium			x		2x

pH = – log[H⁺]

12 = – log [H⁺]

$$ \Rightarrow $$ [H⁺] = 10^–12

As, [H⁺][OH^– ] = 10^–14

10^–12 [OH^– ] = 10^–14

$$ \Rightarrow $$ [OH^– ] = 10^–2

As [OH^– ] = 2x = 10^–2 then x = 5.0 × 10^–3

Now, K_sp = [Ba²⁺][OH^– ]²

K_sp = (5 × 10^–3) (10^–2)² = 5.0 × 10^–7