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Ionic Equilibrium question

2023 · 31 Jan · Shift 2 · Q18
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  5. /2023 · 31 Jan · Shift 2 · Q18

Ionic Equilibrium question

2023 · 31 Jan · Shift 2 · Q18

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
At 298 K298 \mathrm{~K}298 K, the solubility of silver chloride in water is 1.434×10−3 g L−11.434 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}1.434×10−3 g L−1. The value of −log⁡Ksp-\log \mathrm{K}_{\mathrm{sp}}−logKsp​ for silver chloride is ‾\underline{\hspace{2cm}}​. (Given mass of Ag\mathrm{Ag}Ag is 107.9 g mol−1107.9 \mathrm{~g} \mathrm{~mol}^{-1}107.9 g mol−1 and mass of Cl\mathrm{Cl}Cl is 35.5 g mol−135.5 \mathrm{~g} \mathrm{~mol}^{-1}35.5 g mol−1 )
Numerical answer
View written solutionFree

Correct answer: 10

  1. Write the dissolution equilibrium

For silver chloride:

AgCl(s)⇌Ag+(aq)+Cl−(aq)\mathrm{AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)}AgCl(s)⇌Ag+(aq)+Cl−(aq)

If the molar solubility is sss, then

[Ag+]=s,[Cl−]=s[\mathrm{Ag^+}] = s, \qquad [\mathrm{Cl^-}] = s[Ag+]=s,[Cl−]=s

So,

Ksp=[Ag+][Cl−]=s2K_{sp} = [\mathrm{Ag^+}][\mathrm{Cl^-}] = s^2Ksp​=[Ag+][Cl−]=s2

  1. Calculate molar mass of AgCl\mathrm{AgCl}AgCl

Given:

M(Ag)=107.9,M(Cl)=35.5M(\mathrm{Ag}) = 107.9, \qquad M(\mathrm{Cl}) = 35.5M(Ag)=107.9,M(Cl)=35.5

Therefore,

M(AgCl)=107.9+35.5=143.4 g mol−1M(\mathrm{AgCl}) = 107.9 + 35.5 = 143.4\ \mathrm{g\ mol^{-1}}M(AgCl)=107.9+35.5=143.4 g mol−1

  1. Convert solubility from g L−1\mathrm{g\ L^{-1}}g L−1 to mol L−1\mathrm{mol\ L^{-1}}mol L−1

Given solubility:

1.434×10−3 g L−11.434 \times 10^{-3}\ \mathrm{g\ L^{-1}}1.434×10−3 g L−1

Hence,

s=1.434×10−3143.4s = \frac{1.434 \times 10^{-3}}{143.4}s=143.41.434×10−3​

s=1.0×10−5 mol L−1s = 1.0 \times 10^{-5}\ \mathrm{mol\ L^{-1}}s=1.0×10−5 mol L−1

  1. Find KspK_{sp}Ksp​

Ksp=s2=(1.0×10−5)2=1.0×10−10K_{sp} = s^2 = (1.0 \times 10^{-5})^2 = 1.0 \times 10^{-10}Ksp​=s2=(1.0×10−5)2=1.0×10−10

  1. Calculate −log⁡Ksp-\log K_{sp}−logKsp​

−log⁡Ksp=−log⁡(10−10)=10-\log K_{sp} = -\log(10^{-10}) = 10−logKsp​=−log(10−10)=10

  1. Final Answer

10\boxed{10}10​

The derived answer matches the stored correct answer.

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