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Electrochemistry question

2023 · 1 Feb · Shift 2 · Q17
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  5. /2023 · 1 Feb · Shift 2 · Q17

Electrochemistry question

2023 · 1 Feb · Shift 2 · Q17

JEE MainChemistryElectrochemistryNumerical+4 / −1
1×10−5 M AgNO31 \times 10^{-5} ~\mathrm{M} ~\mathrm{AgNO}_{3}1×10−5 M AgNO3​ is added to 1 L1 \mathrm{~L}1 L of saturated solution of AgBr\mathrm{AgBr}AgBr. The conductivity of this solution at 298 K298 \mathrm{~K}298 K is ‾\underline{\hspace{2cm}}​×10−8 S m−1\times 10^{-8} \mathrm{~S} \mathrm{~m}^{-1}×10−8 S m−1. [Given : KSP(AgBr)=4.9×10−13\mathrm{K}_{\mathrm{SP}}(\mathrm{AgBr})=4.9 \times 10^{-13}KSP​(AgBr)=4.9×10−13 at 298 K298 \mathrm{~K}298 KλAg+0=6×10−3 S m2 mol−1λBr−0=8×10−3 S m2 mol−1λNO3−0=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}​λAg+0​=6×10−3 S m2 mol−1λBr−0​=8×10−3 S m2 mol−1λNO3−​0​=7×10−3 S m2 mol−1]​
Numerical answer
View written solutionFree

Correct answer: 13039

  1. Solubility of AgBr in pure water

For saturated AgBr\mathrm{AgBr}AgBr:

AgBr(s)⇌Ag++Br−\mathrm{AgBr(s)} \rightleftharpoons \mathrm{Ag^+} + \mathrm{Br^-}AgBr(s)⇌Ag++Br−

If solubility is sss mol L−1^{-1}−1, then

Ksp=s2=4.9×10−13K_{sp}=s^2=4.9\times 10^{-13}Ksp​=s2=4.9×10−13

So,

s=4.9×10−13=7×10−7 mol L−1s=\sqrt{4.9\times 10^{-13}}=7\times 10^{-7}\ \text{mol L}^{-1}s=4.9×10−13​=7×10−7 mol L−1

Thus initially in 1 L saturated solution:

[Ag+]=[Br−]=7×10−7 M[\mathrm{Ag^+}] = [\mathrm{Br^-}] = 7\times 10^{-7}\ \text{M}[Ag+]=[Br−]=7×10−7 M
  1. Addition of 1×10−51\times 10^{-5}1×10−5 M AgNO3\mathrm{AgNO_3}AgNO3​

AgNO3\mathrm{AgNO_3}AgNO3​ is a strong electrolyte, so it contributes:

[Ag+]from AgNO3=1×10−5 M,[NO3−]=1×10−5 M[\mathrm{Ag^+}]_{\text{from AgNO}_3}=1\times 10^{-5}\ \text{M}, \qquad [\mathrm{NO_3^-}]=1\times 10^{-5}\ \text{M}[Ag+]from AgNO3​​=1×10−5 M,[NO3−​]=1×10−5 M

Because of common ion effect, dissolved AgBr\mathrm{AgBr}AgBr becomes negligible compared to added Ag+\mathrm{Ag^+}Ag+.

Let final dissolved Br−\mathrm{Br^-}Br− concentration be xxx M. Then equilibrium gives:

[Ag+][Br−]=Ksp[\mathrm{Ag^+}][\mathrm{Br^-}] = K_{sp}[Ag+][Br−]=Ksp​

Since added silver ion dominates,

[Ag+]≈1×10−5[\mathrm{Ag^+}] \approx 1\times 10^{-5}[Ag+]≈1×10−5

Hence,

x=[Br−]=4.9×10−131×10−5=4.9×10−8 Mx=[\mathrm{Br^-}] = \frac{4.9\times 10^{-13}}{1\times 10^{-5}}=4.9\times 10^{-8}\ \text{M}x=[Br−]=1×10−54.9×10−13​=4.9×10−8 M

So final concentrations are approximately:

[Ag+]≈1×10−5 M[\mathrm{Ag^+}] \approx 1\times 10^{-5}\ \text{M}[Ag+]≈1×10−5 M [NO3−]=1×10−5 M[\mathrm{NO_3^-}] = 1\times 10^{-5}\ \text{M}[NO3−​]=1×10−5 M [Br−]=4.9×10−8 M[\mathrm{Br^-}] = 4.9\times 10^{-8}\ \text{M}[Br−]=4.9×10−8 M
  1. Convert concentrations to mol m−3^{-3}−3

Since

1 mol L−1=1000 mol m−31\ \text{mol L}^{-1}=1000\ \text{mol m}^{-3}1 mol L−1=1000 mol m−3

we get:

cAg+=1×10−5×1000=1×10−2 mol m−3c_{\mathrm{Ag^+}} = 1\times 10^{-5}\times 1000 = 1\times 10^{-2}\ \text{mol m}^{-3}cAg+​=1×10−5×1000=1×10−2 mol m−3 cNO3−=1×10−2 mol m−3c_{\mathrm{NO_3^-}} = 1\times 10^{-2}\ \text{mol m}^{-3}cNO3−​​=1×10−2 mol m−3 cBr−=4.9×10−8×1000=4.9×10−5 mol m−3c_{\mathrm{Br^-}} = 4.9\times 10^{-8}\times 1000 = 4.9\times 10^{-5}\ \text{mol m}^{-3}cBr−​=4.9×10−8×1000=4.9×10−5 mol m−3
  1. Conductivity formula
κ=∑λi0ci\kappa = \sum \lambda_i^0 c_iκ=∑λi0​ci​

Given:

λAg+0=6×10−3 S m2 mol−1\lambda_{\mathrm{Ag^+}}^0 = 6\times 10^{-3}\ \mathrm{S\, m^2\, mol^{-1}}λAg+0​=6×10−3 Sm2mol−1 λBr−0=8×10−3 S m2 mol−1\lambda_{\mathrm{Br^-}}^0 = 8\times 10^{-3}\ \mathrm{S\, m^2\, mol^{-1}}λBr−0​=8×10−3 Sm2mol−1 λNO3−0=7×10−3 S m2 mol−1\lambda_{\mathrm{NO_3^-}}^0 = 7\times 10^{-3}\ \mathrm{S\, m^2\, mol^{-1}}λNO3−​0​=7×10−3 Sm2mol−1

Therefore,

κ=(6×10−3)(1×10−2)+(7×10−3)(1×10−2)+(8×10−3)(4.9×10−5)\kappa = (6\times 10^{-3})(1\times 10^{-2}) + (7\times 10^{-3})(1\times 10^{-2}) + (8\times 10^{-3})(4.9\times 10^{-5})κ=(6×10−3)(1×10−2)+(7×10−3)(1×10−2)+(8×10−3)(4.9×10−5) κ=6×10−5+7×10−5+3.92×10−7\kappa = 6\times 10^{-5} + 7\times 10^{-5} + 3.92\times 10^{-7}κ=6×10−5+7×10−5+3.92×10−7 κ=1.30392×10−4 S m−1\kappa = 1.30392\times 10^{-4}\ \mathrm{S\, m^{-1}}κ=1.30392×10−4 Sm−1

Now write in the form _____×10−8 S m−1\_\_\_\_\_ \times 10^{-8}\ \mathrm{S\, m^{-1}}_____×10−8 Sm−1:

1.30392×10−4=13039.2×10−81.30392\times 10^{-4} = 13039.2\times 10^{-8}1.30392×10−4=13039.2×10−8

Hence the required integer is

13039\boxed{13039}13039​
  1. Comparison with stored answer

Stored correct answer = 130391303913039

Our derived answer = 130391303913039

So they agree.

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