Hydrofluoric acid is a weak acid. At 25° C, the molar conductivity of 0.002 M HF is 176.2 Ω-1 cm 2 mol -1. If its Lambdam=°=405.2 Ω-1 cm 2 mol -1 . Equilibrium constant at the given concentration is

Question

Hydrofluoric acid is a weak acid. At 25° C, the molar conductivity of 0.002 M HF is 176.2 Ω-1 cm 2 mol -1. If its Lambdam=°=405.2 Ω-1 cm 2 mol -1 . Equilibrium constant at the given concentration is (A) 6.7 × 10-4 M (B) 3.2 × 10-4 M (C) 6.7 × 10-5 M (D) 3.2 × 10-5 M

Tardigrade Admin · Accepted Answer

α=( Lambdac/ Lambdam°)=(176.2 Ω-1 cm 2 mol -1/405.2 Ω-1 cm 2 mol -1) =0.435 K=([ H +][ F -1]/[ HF ])=(C α2/1-α) =((0.002 M )(0.435)2/1-0.435)

Q. Hydrofluoric acid is a weak acid. At $2 5^{\circ} C$ , the molar conductivity of $0.002 M H F$ is $176.2 Ω^{- 1} c m^{2}$ $m o l^{- 1}$ . If its $Λ_{m =}^{\circ} = 405.2 Ω^{- 1} c m^{2} m o l^{- 1} .$ Equilibrium constant at the given concentration is

$6.7 \times 1 0^{- 4} M$

$3.2 \times 1 0^{- 4} M$

$6.7 \times 1 0^{- 5} M$

$3.2 \times 1 0^{- 5} M$

$α = \frac{Λ _{c}}{Λ _{m}^{\circ}} = \frac{176.2 Ω ^{- 1} c m ^{2} m o l ^{- 1}}{405.2 Ω ^{- 1} c m ^{2} m o l ^{- 1}}$

$= 0.435$

$K = \frac{[ H ^{+} ] [ F ^{- 1} ]}{[ H F ]} = \frac{C α ^{2}}{1 - α}$

$= \frac{( 0.002 M ) ( 0.435 ) ^{2}}{1 - 0.435}$

Q. Hydrofluoric acid is a weak acid. At 25∘C, the molar conductivity of 0.002MHF is 176.2Ω−1cm2 mol−1. If its Λm=∘​=405.2Ω−1cm2mol−1. Equilibrium constant at the given concentration is

6.7×10−4M

3.2×10−4M

6.7×10−5M

3.2×10−5M