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ChemHardAdv 2022 • Paper 2

Calculate Sum of Stoichiometric Coefficients from Limiting Molar Conductivity

Consider strong electrolytes ZmXnZ_m X_n, UmYpU_m Y_p, and VmXnV_m X_n. Given that the limiting molar conductivity (Λ0\Lambda^0) of UmYpU_m Y_p and VmXnV_m X_n are 250 S cm2 mol−1250\text{ S cm}^2\text{ mol}^{-1} and 440 S cm2 mol−1440\text{ S cm}^2\text{ mol}^{-1}, respectively. Determine the value of (m+n+p)(m + n + p).

Given:

IonZn+Up+Vn+Xm−Ym−λ0 (S cm2 mol−1)50.025.0100.080.0100.0\begin{array}{|c|c|c|c|c|c|} \hline \text{Ion} & Z^{n+} & U^{p+} & V^{n+} & X^{m-} & Y^{m-} \\ \hline \lambda^0\text{ (S cm}^2\text{ mol}^{-1}\text{)} & 50.0 & 25.0 & 100.0 & 80.0 & 100.0 \\ \hline \end{array}

where λ0\lambda^0 is the limiting molar conductivity of individual ions.

The variation of molar conductivity (Λ\Lambda) of ZmXnZ_m X_n with respect to c1/2c^{1/2} is illustrated in the plot below.

Question Diagram 1
Official Numerical Answer7

Topics & Concepts

ElectrochemistryElectrolysisStoichiometry

Step-by-Step Solution

To determine the value of (m+n+p)(m + n + p), we apply Kohlrausch's law of independent migration of ions to the given strong electrolytes.

1. Extrapolation of Limiting Molar Conductivity of ZmXnZ_m X_n

The variation of molar conductivity (Λ\Lambda) with concentration cc for a strong electrolyte is given by the Debye-Hückel-Onsager equation: Λ=Λ0−bc\Lambda = \Lambda^0 - b \sqrt{c}

From the provided plot of Λ\Lambda vs c1/2c^{1/2} for ZmXnZ_m X_n:

  • At c1/2=0.01 (mol L−1)1/2c^{1/2} = 0.01 \text{ (mol L}^{-1}\text{)}^{1/2}, Λ=339 S cm2 mol−1\Lambda = 339 \text{ S cm}^2 \text{ mol}^{-1}
  • At c1/2=0.04 (mol L−1)1/2c^{1/2} = 0.04 \text{ (mol L}^{-1}\text{)}^{1/2}, Λ=336 S cm2 mol−1\Lambda = 336 \text{ S cm}^2 \text{ mol}^{-1}

The slope bb is calculated as: b=−ΔΛΔ(c1/2)=−336−3390.04−0.01=100 S cm2 mol−1(mol L−1)−1/2b = -\frac{\Delta \Lambda}{\Delta (c^{1/2})} = -\frac{336 - 339}{0.04 - 0.01} = 100 \text{ S cm}^2 \text{ mol}^{-1} \text{(mol L}^{-1}\text{)}^{-1/2}

The limiting molar conductivity (Λ0\Lambda^0) of ZmXnZ_m X_n is the y-intercept at c1/2=0c^{1/2} = 0: Λ0(ZmXn)=339+100×0.01=340 S cm2 mol−1\Lambda^0(Z_m X_n) = 339 + 100 \times 0.01 = 340 \text{ S cm}^2 \text{ mol}^{-1}

2. Formulating Equations using Kohlrausch's Law

By Kohlrausch's law, the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its constituent ions multiplied by their respective stoichiometric coefficients:

  1. For UmYpU_m Y_p: Λ0(UmYp)=m⋅λ0(Up+)+p⋅λ0(Ym−)\Lambda^0(U_m Y_p) = m \cdot \lambda^0(U^{p+}) + p \cdot \lambda^0(Y^{m-}) 250=m(25.0)+p(100.0)250 = m(25.0) + p(100.0) Dividing the equation by 2525: m+4p=10— (Equation 1)m + 4p = 10 \quad \text{--- (Equation 1)}

  2. For VmXnV_m X_n: Λ0(VmXn)=m⋅λ0(Vn+)+n⋅λ0(Xm−)\Lambda^0(V_m X_n) = m \cdot \lambda^0(V^{n+}) + n \cdot \lambda^0(X^{m-}) 440=m(100.0)+n(80.0)440 = m(100.0) + n(80.0) Dividing the equation by 2020: 5m+4n=22— (Equation 2)5m + 4n = 22 \quad \text{--- (Equation 2)}

  3. For ZmXnZ_m X_n: Λ0(ZmXn)=m⋅λ0(Zn+)+n⋅λ0(Xm−)\Lambda^0(Z_m X_n) = m \cdot \lambda^0(Z^{n+}) + n \cdot \lambda^0(X^{m-}) 340=m(50.0)+n(80.0)340 = m(50.0) + n(80.0) Dividing the equation by 1010: 5m+8n=34— (Equation 3)5m + 8n = 34 \quad \text{--- (Equation 3)}

3. Solving for m,n,m, n, and pp

Subtracting Equation 2 from Equation 3: (5m+8n)−(5m+4n)=34−22(5m + 8n) - (5m + 4n) = 34 - 22 4n=12  ⟹  n=34n = 12 \implies n = 3

Substituting n=3n = 3 into Equation 2: 5m+4(3)=225m + 4(3) = 22 5m=10  ⟹  m=25m = 10 \implies m = 2

Substituting m=2m = 2 into Equation 1: 2+4p=102 + 4p = 10 4p=8  ⟹  p=24p = 8 \implies p = 2

4. Calculating (m+n+p)(m + n + p)

m+n+p=2+3+2=7m + n + p = 2 + 3 + 2 = 7

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Calculate Sum of Stoichiometric Coefficients from Limiting Molar Conductivity | Chemistry PYQ Solution - JEE Challenger