Given that i n S c m 2 e q - 1 at T = 298 K : Λ e q o for B a O H 2 , B a C l 2 and N H 4 C l are 228.8 , 120.3 and 129.8 respectively. Specific conductance for 0.2 N N H 4 O H solution is 4.766 × 10 - 4 S c m - 1 , then value of p H of the given solution of N H 4 O H will be nearly. (Take log ⁡ 2 = 0.3 )

Step-by-step solution

Λ e q o B a O H 2 = λ o B a 2 + + λ e q o O H - .....(I) λ e q o B a C l 2 = λ e q o B a 2 + λ e q o C l - .....(II) λ e q o N H 4 C l = λ e q o N H 4 + + λ e q o C l - .....(III) λ e q o N H 4 O H = λ e q o N H 4 + + λ e q o O H - I + I II - I I λ e q o N H 4 O H = 228.8 + 129.8 - 120.3 = 238.33 c m 2 e q - 1 λ e q N H 4 O H = 4.766 × 10 - 4 × 1000 0.2 = 2.383 α = λ e q N H 4 O H λ e q 1 o N H 4 O H = 10 - 2 N H 4 O H c 1 - α ⇌ N H 4 + c α + O H - c α O H - = 0.2 × 10 - 2 = 2 × 10 - 3 p O H = 3 - log ⁡ 2 ⇒ p H = 1