JEE Main Chemistry question
The solubility of PbCl₂ in water is 0.01 M at 25°C. Its maximum concentration in 0.1 M NaCl will be:
Answer: (D)
Solution
Ksp=4(0.01)³=4×10⁻⁶. In 0.1 M NaCl: [Pb²⁺]=Ksp/[Cl⁻]²=4×10⁻⁶/0.01=4×10⁻⁴ M.
The solubility of PbCl₂ in water is 0.01 M at 25°C. Its maximum concentration in 0.1 M NaCl will be:
Ksp=4(0.01)³=4×10⁻⁶. In 0.1 M NaCl: [Pb²⁺]=Ksp/[Cl⁻]²=4×10⁻⁶/0.01=4×10⁻⁴ M.