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Equilibrium: JEE Main Chemistry Question with Solution At
298 K 298\mathrm{~K} 298 K , the solubility of silver chloride in water is
1.434 × 10 − 3 g L − 1 1.434 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1} 1.434 × 1 0 − 3 g L − 1 . The value of
− log K s p -\log \mathrm{K}_{\mathrm{sp}} − log K sp for silver chloride is
( Given mass of A g is 107.9 g m o l − 1 and mass of C l is 35.5 g m o l − 1 ) (\text{Given mass of } \mathrm{Ag} \text{ is } 107.9\mathrm{~g} \mathrm{~mol}^{-1} \text{ and mass of } \mathrm{Cl} \text{ is } 35.5\mathrm{~g}\mathrm{~mol}^{-1}) ( Given mass of Ag is 107.9 g mol − 1 and mass of Cl is 35.5 g mol − 1 ) Hint 1 of 3
What is the molar mass of A g C l \mathrm{AgCl} AgCl
143.4 g mol − 1 143.4\text{ g mol}^{-1} 143.4 g mol − 1 142.4 g mol − 1 142.4\text{ g mol}^{-1} 142.4 g mol − 1 No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.
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The value of − log K sp -\log K_{\text{sp}} − log K sp 10 10 10 Option analysis
Why each option works or fails
Step Working
01 given Solubility of A g C l \mathrm{AgCl} AgCl s = 1.434 × 10 − 3 g L − 1 s = 1.434 \times 10^{-3}\text{ g L}^{-1} s = 1.434 × 1 0 − 3 g L − 1 A g = 107.9 g mol − 1 \mathrm{Ag} = 107.9\text{ g mol}^{-1} Ag = 107.9 g mol − 1 C l = 35.5 g mol − 1 \mathrm{Cl} = 35.5\text{ g mol}^{-1} Cl = 35.5 g mol − 1 T = 298 K T = 298\text{ K} T = 298 K
02 find The value of − log K s p -\log K_{\mathrm{sp}} − log K sp
03 strategise 1. Compute the molar mass of A g C l \mathrm{AgCl} AgCl M = M A g + M C l M = M_{\mathrm{Ag}} + M_{\mathrm{Cl}} M = M Ag + M Cl g L − 1 \text{g L}^{-1} g L − 1 S (mol L − 1 ) S\text{ (mol L}^{-1}\text{)} S (mol L − 1 ) S = s / M S = s / M S = s / M A g C l ( s ) ⇌ A g + ( a q ) + C l − ( a q ) \mathrm{AgCl(s)} \rightleftharpoons \mathrm{Ag^+}(aq) + \mathrm{Cl^-}(aq) AgCl ( s ) ⇌ A g + ( a q ) + C l − ( a q ) K s p = S 2 K_{\mathrm{sp}} = S^2 K sp = S 2 − log K s p = − log ( S 2 ) = − 2 log S -\log K_{\mathrm{sp}} = -\log(S^2) = -2\log S − log K sp = − log ( S 2 ) = − 2 log S
04 execute Calculate molar mass: M = 107.9 + 35.5 = 143.4 g mol − 1 M = 107.9 + 35.5 = 143.4\text{ g mol}^{-1} M = 107.9 + 35.5 = 143.4 g mol − 1 S = 1.434 × 10 − 3 143.4 = 1.0 × 10 − 5 mol L − 1 S = \frac{1.434 \times 10^{-3}}{143.4} = 1.0 \times 10^{-5}\text{ mol L}^{-1} S = 143.4 1.434 × 1 0 − 3 = 1.0 × 1 0 − 5 mol L − 1 K s p = S 2 = ( 1.0 × 10 − 5 ) 2 = 1.0 × 10 − 10 K_{\mathrm{sp}} = S^2 = (1.0 \times 10^{-5})^2 = 1.0 \times 10^{-10} K sp = S 2 = ( 1.0 × 1 0 − 5 ) 2 = 1.0 × 1 0 − 10 − log K s p = − log ( 10 − 10 ) = 10 -\log K_{\mathrm{sp}} = -\log(10^{-10}) = 10 − log K sp = − log ( 1 0 − 10 ) = 10
✓ verify 1.434 / 143.4 = 10 − 2 1.434 / 143.4 = 10^{-2} 1.434/143.4 = 1 0 − 2 10 − 3 10^{-3} 1 0 − 3 10 − 5 M 10^{-5}\text{ M} 1 0 − 5 M 10 − 10 10^{-10} 1 0 − 10 − log ( 10 − 10 ) = 10 -\log(10^{-10}) = 10 − log ( 1 0 − 10 ) = 10
Hints that build this answer step by step What is the molar mass of A g C l \mathrm{AgCl} AgCl
143.4 g mol − 1 143.4\text{ g mol}^{-1} 143.4 g mol − 1 What is the molar solubility (s s s A g C l \mathrm{AgCl} AgCl mol L − 1 \text{mol L}^{-1} mol L − 1
1.0 × 10 − 5 mol L − 1 1.0 \times 10^{-5}\text{ mol L}^{-1} 1.0 × 1 0 − 5 mol L − 1 Using K sp = s 2 K_{\text{sp}} = s^2 K sp = s 2 − log K sp -\log K_{\text{sp}} − log K sp
10 10 10 Your next move We think you should solve this next ✓ Source and academic review↓
Question type Numerical
Exam relevance JEE Main · Chemistry
Concepts assessed Chemistry
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Why is Ksp = S^2 instead of 4S^3? AgCl is a 1:1 salt (AB type), dissociating into one Ag+ and one Cl- ion. Hence, Ksp = [Ag+][Cl-] = (S)(S) = S^2.
Answer The value of − log K sp -\log K_{\text{sp}} − log K sp 10 10 10
Why each option works or fails Step-by-step solution given: Solubility of A g C l \mathrm{AgCl} AgCl s = 1.434 × 10 − 3 g L − 1 s = 1.434 \times 10^{-3}\text{ g L}^{-1} s = 1.434 × 1 0 − 3 g L − 1 A g = 107.9 g mol − 1 \mathrm{Ag} = 107.9\text{ g mol}^{-1} Ag = 107.9 g mol − 1 C l = 35.5 g mol − 1 \mathrm{Cl} = 35.5\text{ g mol}^{-1} Cl = 35.5 g mol − 1 T = 298 K T = 298\text{ K} T = 298 K find: The value of − log K s p -\log K_{\mathrm{sp}} − log K sp strategise: 1. Compute the molar mass of A g C l \mathrm{AgCl} AgCl M = M A g + M C l M = M_{\mathrm{Ag}} + M_{\mathrm{Cl}} M = M Ag + M Cl g L − 1 \text{g L}^{-1} g L − 1 S (mol L − 1 ) S\text{ (mol L}^{-1}\text{)} S (mol L − 1 ) S = s / M S = s / M S = s / M A g C l ( s ) ⇌ A g + ( a q ) + C l − ( a q ) \mathrm{AgCl(s)} \rightleftharpoons \mathrm{Ag^+}(aq) + \mathrm{Cl^-}(aq) AgCl ( s ) ⇌ A g + ( a q ) + C l − ( a q ) K s p = S 2 K_{\mathrm{sp}} = S^2 K sp = S 2 − log K s p = − log ( S 2 ) = − 2 log S -\log K_{\mathrm{sp}} = -\log(S^2) = -2\log S − log K sp = − log ( S 2 ) = − 2 log S execute: Calculate molar mass: M = 107.9 + 35.5 = 143.4 g mol − 1 M = 107.9 + 35.5 = 143.4\text{ g mol}^{-1} M = 107.9 + 35.5 = 143.4 g mol − 1 S = 1.434 × 10 − 3 143.4 = 1.0 × 10 − 5 mol L − 1 S = \frac{1.434 \times 10^{-3}}{143.4} = 1.0 \times 10^{-5}\text{ mol L}^{-1} S = 143.4 1.434 × 1 0 − 3 = 1.0 × 1 0 − 5 mol L − 1 K s p = S 2 = ( 1.0 × 10 − 5 ) 2 = 1.0 × 10 − 10 K_{\mathrm{sp}} = S^2 = (1.0 \times 10^{-5})^2 = 1.0 \times 10^{-10} K sp = S 2 = ( 1.0 × 1 0 − 5 ) 2 = 1.0 × 1 0 − 10 − log K s p = − log ( 10 − 10 ) = 10 -\log K_{\mathrm{sp}} = -\log(10^{-10}) = 10 − log K sp = − log ( 1 0 − 10 ) = 10 verify: 1.434 / 143.4 = 10 − 2 1.434 / 143.4 = 10^{-2} 1.434/143.4 = 1 0 − 2 10 − 3 10^{-3} 1 0 − 3 10 − 5 M 10^{-5}\text{ M} 1 0 − 5 M 10 − 10 10^{-10} 1 0 − 10 − log ( 10 − 10 ) = 10 -\log(10^{-10}) = 10 − log ( 1 0 − 10 ) = 10