NH₄OH and NH₄Cl constitute to form a basic buffer.

pOH = pK_b + log $${{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$$

We know, pOH+ pH = 14 or pOH = 14 – pH

$$ \therefore $$ 14 - pH - $$\log {{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$$ = pK_b

$$ \Rightarrow $$ 14 - 9.25 - $$\log {{0.1} \over {0.1}}$$ = pK_b

$$ \Rightarrow $$ 14 – 9.25 – 0 = pK_b

$$ \Rightarrow $$ pK_b = 4.75

Na₂CO₃ is a salt of weak acid H₂CO₃ and strong base NaOH, therefore, its aqueous solution will be basic hence has pH more than 7.

For the reaction

BaO_2(g) $$\rightleftharpoons$$ BaO_(s) + O_2(g); H = +ve.

At equilibrium K_p = P_O2

[For solid and liquids concentration term is taken as unity]

Hence, the value of equilibrium constant depends only upon partial pressure of O₂. Further on increasing temperature formation of O₂ increases as this is an endothermic reaction.

MX2	⇌	Ag2+	+	2X-
s		s		2s

K_sp = [M²⁺] [X^–]² = (S)(2S)² = 4S³

$$ \Rightarrow $$ K_sp = 4(0.5 × 10^–4)³ = 5 × 10^–13