Question

At 298 K , the conductivity of a saturated solution of AgCl in water is 2.6 × 10^–6 ohm^–1 cm^–1. Given ,$\lambda_{m}^{\infty}$ (Ag⁺) = 63 ohm^–1 cm²mol^–1&$\lambda_{m}^{\infty}$ (Cl^–) = 67 ohm^–1 cm²mol^–1

Therefore solubility product of AgCl is

• A. 2 × 10^–5
• B. 4 × 10^–10
• C. 4 × 10^–16
• D. 2 × 10^–8

Answer 1

Answer: (B)

4 × 10^–10

Answer 2

s = $\frac{10^{3} \times \kappa}{\lambda_{m}^{\infty}(AgCl)}$ = $\frac{10^{3} \times 2.6 \times 10^{–6}}{(63 + 67)}$= 2 × 10^–5 (M)

\ K_sp = ( 2 × 10^–5)² = 4 × 10^–10.