At 298 K, the conductivity of a saturated solution of AgCl in water is 2.6 × 10-6 ohm -1 cm -1. Given λm∞( Ag +)=63 ohm -1 cm 2 mol -1 and λm∞( Cl -)=67 ohm -1 cm 2 mol -1 Therefore solubility product of AgCl is

Question

At 298 K, the conductivity of a saturated solution of AgCl in water is 2.6 × 10-6 ohm -1 cm -1. Given λm∞( Ag +)=63 ohm -1 cm 2 mol -1 and λm∞( Cl -)=67 ohm -1 cm 2 mol -1 Therefore solubility product of AgCl is (A) 2 × 10-5 (B) 4 × 10-10 (C) 4 × 10-16 (D) 2 × 10-8

Tardigrade Admin · Accepted Answer

s=(103 × kappa/λm∞( AgCl ))=(103 × 2.6 × 10-6/(63+67)) =2 × 10-5( M ) ∴ K sp =(2 × 10-5)2=4 × 10-10

$2 \times 1 0^{- 5}$

$4 \times 1 0^{- 10}$

$4 \times 1 0^{- 16}$

$2 \times 1 0^{- 8}$

$s = \frac{1 0 ^{3} \times κ}{λ _{m}^{\infty} ( A g Cl )} = \frac{1 0 ^{3} \times 2.6 \times 1 0 ^{- 6}}{( 63 + 67 )}$
$= 2 \times 1 0^{- 5} (M)$
$∴ K_{s p} = (2 \times 1 0^{- 5})^{2} = 4 \times 1 0^{- 10}$

Q. At 298K, the conductivity of a saturated solution of AgCl in water is 2.6×10−6ohm−1cm−1. Given λm∞​(Ag+)=63ohm−1cm2mol−1 and λm∞​(Cl−)=67ohm−1cm2mol−1 Therefore solubility product of AgCl is

2×10−5

4×10−10

4×10−16

2×10−8

s=λm∞​(AgCl)103×κ​=(63+67)103×2.6×10−6​ =2×10−5(M) ∴Ksp​=(2×10−5)2=4×10−10

$2 \times 1 0^{- 5}$

$4 \times 1 0^{- 10}$

$4 \times 1 0^{- 16}$

$2 \times 1 0^{- 8}$

$s = \frac{1 0 ^{3} \times κ}{λ _{m}^{\infty} ( A g Cl )} = \frac{1 0 ^{3} \times 2.6 \times 1 0 ^{- 6}}{( 63 + 67 )}$
$= 2 \times 1 0^{- 5} (M)$
$∴ K_{s p} = (2 \times 1 0^{- 5})^{2} = 4 \times 1 0^{- 10}$