Published by:
CGP EDU Academic Team
Published on: August 14, 2026
The minimum volume of water required to dissolve 0.1 g lead(II) chloride to get a saturated solution (K sp of PbCl 2 = 3.2 × 10 –8 ; atomic mass of Pb = 207 u) is :
Text Solution
Verified by ExpertsThe correct answer is:
D
= 32 × 10 –9
PbCl 2 
Pb 2+ + 2Cl –
s 2s
K sp = [Pb 2+ ][Cl – ] 2
K sp = 4s 3 = 32 × 10 –9
s 3 = 8 × 10 – 9
s = 2 × 10 –3 M
× 
= 2 × 10 –3
× 
= 2 × 10 –3
V L = 
= 0.18 L
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, K a of th…
20 mL of 0.1 M H 2 SO 4 solution is added to 30 mL of 0.2 M NH 4 OH solution. The pH of the resulta…
An aqueous solution contains an unknown concentration of Ba 2+ . When 50 mL of a 1M solution of Na …
A mixture of 100 m mol of Ca(OH) 2 and 2 g of sodium sulphate was dissolved in water and the volume…
A weak acid HX at 25ºC has the dissociation constant 1 × 10 –5 . It forms a salt NaX on reaction wi…
The dissociation constant of a substituted benzoic acid at 25ºC is 1.0 × 10 –4 . The pH of 0.01 M s…