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Q.

Following data is given at  T = 298 K λ∘ eq for Ba(OH)2=228.8ohm−1cm2eq−1λ∘ eq for BaCℓ2=120.3ohm−1cm2eq−1λ0 eq for NH4Cℓ=129.8ohm−1cm2eq−1Specific conductance for 0.2N    NH4OH solution is4.766×10–4ohm–1cm–1Then the value of pH of the given solution of NH4OH will be nearly.

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a

9.2

b

11.3

c

12.1

d

2.7

answer is B.

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Detailed Solution

λBa(OH)2∘=λBa+2∘+λOH−∘λBaCl2∘=λBa+2∘+λCl−∘λNH4Cl∘=λNH4∘+λCl−∘⇒λNH4OH∘=λNH4∘+λOH−∘=129.8+222.8−120.3=238.3∧m∘λ=K×1000C=4.766×10−4×10000.2⇒α=∧∧∘=0.01⇒OH−=0.01×0.2=2×10−3⇒POH−3−log⁡2pH = 11.3
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