NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
Calculate the volume of water(in mL) required to dissolve 0.1 g lead (II) chloride to get a saturated solution (K sp of PbCl 2 = 3.21 × 10 -8 , atomic mass of Pb = 207 u).Report your answer by rounding it up to nearest whole number.
Correct answer
180
Step-by-step solution
Suppose solubility of PbCl 2 in water is s mol L -1 . PbCl 2 s ⇌ Pb 2 + (aq) + 2Cl - (aq) 1 - s s 2s K sp = [Pb 2 + ] · [Cl - ] 2 K sp = [s] [2s] 2 = 4s 3 3.2 × 10 -8 = 4s 3 s 3 = 3.2 × 10 - 8 4 = 0.8 × 10 - 8 s 3 = 8.0 × 10 -9 Solubility of PbCl 2 , = s = 2 × 10 -3 mol L -1 Solubility of PbCl 2 in gL -1 = 278 × 2 × 10 -3 = 0.556 g L -1 (∵ Molar mass of PbCl 2 = 207 + (2 × 35.5) = 278) 0.556 g of PbCl 2 dissolve in 1 L of water. ∴ 0.1 g of PbCl 2 will dissolve in = 1 × 0.1 0.556 L of water = 0.1798 L = 179.8 mL