allerySolubility product Ksp of AgCl is 1.6 × 10⁻¹⁰. Its solubility in pure water (mol/L):
A8 × 10⁻¹⁰
B1.26 × 10⁻⁵
C1.6 × 10⁻⁵
D1.6 × 10⁻¹⁰
Answer & Solution
Correct answer: B. 1.26 × 10⁻⁵
AgCl ⇌ Ag+ + Cl-.
If solubility = s, then [Ag+][Cl-] = s² = Ksp.
We have s = √(1.6 × 10⁻¹⁰).
So s = 1.26 × 10⁻⁵ mol/L.
Keep practising Unit 7A1: Equilibrium Constant and Dynamic Equilibrium
Browse all Unit 7A1: Equilibrium Constant and Dynamic Equilibrium questions for JEE Advanced, or explore more Chemistry topics.
Related questions
Degree of dissociation α of weak acid HA (Ka = 1.8 × 10⁻⁵) in 0.01 M solution:For the reaction H2(g) + I2(g) ⇌ 2HI(g) with Kc = 50 at 800 K, if [H2] = [I2] = 0.01 M iniIn a saturated solution of AgCl, [Ag+] = 1.26 × 10⁻⁵. If KNO3 is added (no common ion), soAt equilibrium, the ratio of forward to reverse reaction rates:At 298 K, water has Kw = 10⁻¹⁴. Concentration of H+ in pure water:If K = 1 for a reaction, the standard free energy change ΔG°:For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), at equilibrium increasing pressure:For the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), if initial pressure of PCl5 is P0 and degree