The limiting molar conductivities at 2 5 ∘ C of HCl , NaCl and NaA are 426 Ω − 1 cm 2 mole − 1 ; 126 Ω − 1 cm 2 mole − 1 and 83 Ω − 1 cm 2 mole − 1 respectively. The conductivity of 0.001 M acid (HA)

Question

The limiting molar conductivities at 2 5 ∘ C of HCl , NaCl and NaA are 426 Ω − 1 cm 2 mole − 1 ; 126 Ω − 1 cm 2 mole − 1 and 83 Ω − 1 cm 2 mole − 1 respectively. The conductivity of 0.001 M acid (HA) solution is 3.83 × 1 0 − 5 Ω − 1 c m − 1 . The pH of the acid solution is

TestHub · Accepted Answer

λ m ∞ ​ ( H A ) = λ m ∞ ​ ( N a A ) + λ m ∞ ​ ( H Cl ) − λ m ∞ ​ ( N a Cl ) = 83 + 426 − 126 = 283 Ω − 1 cm 2 mole − 1 λ m c ​ = m k × 1000 ​ = 1 0 − 3 3.83 × 1 0 − 5 × 1000 ​ = 38.3 Degree of dissociation, α = λ m ∞ ​ λ m c ​ ​ = 383 38.3 ​ = 0.1 [ H + ] = c α = 0.1 × 1 0 − 3 = 1 0 − 4 ; p H = 4

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