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NVAMediumJEE 2023Electrolytic Conductance & Kohlrausch's Law

JEE Chemistry 2023 Question with Solution

1×10−51 \times 10^{-5} M AgNO3_3 is added to 1 L1 \text{ L} of saturated solution of AgBr. The conductivity of this solution at 298 K298 \text{ K} is _____ ×10−8\times 10^{-8} S m−1{}^{-1}.

Given:

Ksp_{sp}(AgBr) = 4.9×10−134.9 \times 10^{-13} at 298K298\text{K}

λAg+0\lambda^0_{Ag^+} = 6×10−36 \times 10^{-3} Sm2{}^{2} mol−1{}^{-1}

λBr−0\lambda^0_{Br^-} = 8×10−38 \times 10^{-3} Sm2{}^{2} mol−1{}^{-1}

λNO3−0\lambda^0_{NO_3^-} = 7×10−37 \times 10^{-3} Sm2{}^{2} mol−1{}^{-1}

Answer

Correct answer:14

Step-by-step solution

Standard Method

Given: 1×10−5 M1 \times 10^{-5} \text{ M} AgNO3_3 is added to 1 L1 \text{ L} of saturated AgBr solution. Ksp(AgBr)=4.9×10−13K_{\text{sp}}(\text{AgBr}) = 4.9 \times 10^{-13}, λAg+0=6×10−3 S m2 mol−1\lambda^0_{\text{Ag}^+} = 6 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}, λBr−0=8×10−3 S m2 mol−1\lambda^0_{\text{Br}^-} = 8 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}, and λNO3−0=7×10−3 S m2 mol−1\lambda^0_{\text{NO}_3^-} = 7 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}.

Find: The conductivity of the resulting solution.

The dissociation of AgBr is:

AgBr(s)⇋Ag+(aq)+Br−(aq)\text{AgBr}(s) \leftrightharpoons \text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) Ksp=[Ag+][Br−]=4.9×10−13K_{\text{sp}} = [\text{Ag}^+][\text{Br}^-] = 4.9 \times 10^{-13}

Let the solubility of AgBr be s mol/Ls \text{ mol/L}. Then,

[Ag+]=s,[Br−]=s[\text{Ag}^+] = s, \qquad [\text{Br}^-] = s

So,

s2=4.9×10−13s^2 = 4.9 \times 10^{-13} s=4.9×10−13=7×10−7 Ms = \sqrt{4.9 \times 10^{-13}} = 7 \times 10^{-7} \text{ M}

Thus, in saturated AgBr solution,

[Ag+]=[Br−]=7×10−7 M[\text{Ag}^+] = [\text{Br}^-] = 7 \times 10^{-7} \text{ M}

After adding 1×10−5 M1 \times 10^{-5} \text{ M} AgNO3_3, the concentration of Ag+\text{Ag}^+ from AgNO3_3 is much larger than that from AgBr, so

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

Also,

[NO3−]=1×10−5 M[\text{NO}_3^-] = 1 \times 10^{-5} \text{ M}

and the solution uses

[Br−]≈7×10−7 M[\text{Br}^-] \approx 7 \times 10^{-7} \text{ M}

Now use the conductivity relation:

κ=∑λici\kappa = \sum \lambda_i c_i

Therefore,

κ=λAg+[Ag+]+λBr−[Br−]+λNO3−[NO3−]\kappa = \lambda_{\text{Ag}^+}[\text{Ag}^+] + \lambda_{\text{Br}^-}[\text{Br}^-] + \lambda_{\text{NO}_3^-}[\text{NO}_3^-]

Substituting the values:

κ=(6×10−3)(1×10−5)+(8×10−3)(7×10−7)+(7×10−3)(1×10−5)\kappa = (6 \times 10^{-3})(1 \times 10^{-5}) + (8 \times 10^{-3})(7 \times 10^{-7}) + (7 \times 10^{-3})(1 \times 10^{-5}) κ=6×10−8+5.6×10−9+7×10−8\kappa = 6 \times 10^{-8} + 5.6 \times 10^{-9} + 7 \times 10^{-8} κ≈13.56×10−8=14×10−8 S m−1\kappa \approx 13.56 \times 10^{-8} = 14 \times 10^{-8} \text{ S m}^{-1}

Therefore, the conductivity is 14×10−8 S m−114 \times 10^{-8} \text{ S m}^{-1}.

Using solubility and ionic contribution

Given: A saturated AgBr solution and added AgNO3_3 of concentration 1×10−5 M1 \times 10^{-5} \text{ M}.

Find: The total conductivity from all ions present.

First calculate the solubility of AgBr from its solubility product:

Ksp=s2=4.9×10−13K_{\text{sp}} = s^2 = 4.9 \times 10^{-13} s=7×10−7 Ms = 7 \times 10^{-7} \text{ M}

So the saturated AgBr solution initially contains Ag+\text{Ag}^+ and Br−\text{Br}^- each at 7×10−7 M7 \times 10^{-7} \text{ M}.

On adding AgNO3_3, the ions contributed are mainly Ag+\text{Ag}^+ and NO3−\text{NO}_3^-, each of concentration 1×10−5 M1 \times 10^{-5} \text{ M}. The conductivity is obtained by summing the individual ionic conductance contributions.

The three contributions are:

λAg+[Ag+]=(6×10−3)(1×10−5)=6×10−8\lambda_{\text{Ag}^+}[\text{Ag}^+] = (6 \times 10^{-3})(1 \times 10^{-5}) = 6 \times 10^{-8} λBr−[Br−]=(8×10−3)(7×10−7)=5.6×10−9\lambda_{\text{Br}^-}[\text{Br}^-] = (8 \times 10^{-3})(7 \times 10^{-7}) = 5.6 \times 10^{-9} λNO3−[NO3−]=(7×10−3)(1×10−5)=7×10−8\lambda_{\text{NO}_3^-}[\text{NO}_3^-] = (7 \times 10^{-3})(1 \times 10^{-5}) = 7 \times 10^{-8}

Adding them,

κ=6×10−8+5.6×10−9+7×10−8=13.56×10−8 S m−1\kappa = 6 \times 10^{-8} + 5.6 \times 10^{-9} + 7 \times 10^{-8} = 13.56 \times 10^{-8} \text{ S m}^{-1}

Rounding to the asked numerical value gives 1414.

Therefore, the required answer is 14.

Common mistakes

  • Using only the ions from AgNO3_3 and ignoring the ions already present in saturated AgBr is incorrect, because the dissolved AgBr still contributes to conductivity. Include the Br−\text{Br}^- contribution as shown in the solution.

  • Confusing solubility ss with KspK_{\text{sp}} is incorrect. For AgBr, Ksp=s2K_{\text{sp}} = s^2, so the ion concentration from saturation is s=Ksps = \sqrt{K_{\text{sp}}}, not KspK_{\text{sp}} itself.

  • Forgetting that conductivity is the sum of ionic contributions is incorrect. Do not use a single molar conductivity value for the whole mixture; instead apply κ=∑λici\kappa = \sum \lambda_i c_i for all ions present.

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