Question

please ans it fast very important
class 11th chemisty ​

Answer 1

2 × 10⁻⁶ M

Explanation:

Ksp/Ka = Ka

Ka = [AgCN] / [HCN]

    = 2.48 × 10⁻¹⁶ / 6.2 × 10⁻¹⁰

    = 4 × 10⁻⁷  

At equillibrium,

AgCN + H⁺ ⇄ HCN + Ag⁺

AgCN = x moles  

[H⁺] = 10⁻⁵ At pH = 5

Ka = [Ag⁺] [HCN⁺]/[H⁺]

4 × 10⁻⁷ =  x²/  10⁻⁵

x² = 4 × 10⁻⁷ × 10⁻⁵

   = 4 × 10⁻¹²

   = 2 × 10⁻⁶ M

Molar solubility of the AgCN in buffer solution is 2 × 10⁻⁶ M