Verified Solution • 2025–26 Session moderate

QJEEM23Ch26b6: JEE Main 2023 Chemistry - Class 12 JEEM23Ch26b6 Solution: is added to of saturated solution of . The conductivity of this …

1×10−5 M AgNO31 \times 10^{-5} ~\mathrm{M} ~\mathrm{AgNO}_{3} is added to 1 L1 \mathrm{~L} of saturated solution of AgBr\mathrm{AgBr} . The conductivity of this solution at 298 K298 \mathrm{~K} is _____________ ×10−8 S m−1\times 10^{-8} \mathrm{~S} \mathrm{~m}^{-1} .

[Given : KSP(AgBr)=4.9×10−13\mathrm{K}_{\mathrm{SP}}(\mathrm{AgBr})=4.9 \times 10^{-13} at 298 K298 \mathrm{~K}

amp;λAg+0=6×10−3 S m2 mol−1amp;λBr−0=8×10−3 S m2 mol−1amp;λNO3−0=7×10−3 S m2 mol−1]\begin{aligned}& \lambda_{\mathrm{Ag}^{+}}^{0}=6 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\& \lambda_{\mathrm{Br}^{-}}^{0}=8 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\& \left.\lambda_{\mathrm{NO}_{3}^{-}}^{0}=7 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\right]\end{aligned}

Step-by-Step Verified Solution

amp;AgBr⁡(S)⇌⇌Ag+(10−5+x)(aq)+Brx−(aq)amp;x(x+10−5)=4.9×10−13amp;x≃4.9×10−8Mamp;λAg+∘=6×10−3 S cm2 mol−1amp;λBr−0=8×10−3 S cm2 mol−1amp; Ksolution =KAg++KBr−+KNO3−amp;=6×10−3×10−5×103+8×10−3×4.9×10−8×103amp;+7×10−3×10−5×103amp;=(6000+39.2+7000)×10−8amp;=13039.2×10−8 Sm−1\begin{aligned}& \operatorname{AgBr}(\mathrm{S}) \rightleftharpoons \underset{\left(10^{-5}+\mathrm{x}\right)}{\rightleftharpoons \mathrm{Ag}^{+}}(\mathrm{aq})+\mathrm{Br}_{\mathrm{x}}^{-}(\mathrm{aq}) \\\\& x\left(x+10^{-5}\right)=4.9 \times 10^{-13} \\\\& x \simeq 4.9 \times 10^{-8} \mathrm{M} \\\\& \lambda_{\mathrm{Ag}^{+}}^{\circ}=6 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\& \lambda_{\mathrm{Br}^{-}}^0=8 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\& \mathrm{~K}_{\text {solution }}=\mathrm{K}_{\mathrm{Ag}^{+}}+\mathrm{K}_{\mathrm{Br}^{-}}+\mathrm{K}_{\mathrm{NO}_3^{-}} \\\\& =6 \times 10^{-3} \times 10^{-5} \times 10^3+8 \times 10^{-3} \times 4.9 \times 10^{-8} \times 10^3 \\\\& +7 \times 10^{-3} \times 10^{-5} \times 10^3 \\\\& =(6000+39.2+7000) \times 10^{-8} \\\\& =13039.2 \times 10^{-8} ~\mathrm{Sm}^{-1}\end{aligned}

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