For the strong electrolytes NaOH, NaCl and $BaCl_{2}$ the molar conductances at infinite dilution are $248.1 imes10^{-4}$, $126.5 imes10^{-4}$ and $280 imes10^{-4}$ $sm^{2}mol^{-1}$ respectively, then $\lambda_{m}^{\circ}$ for $Ba(OH)_{2}$ in $sm^{2}mol^{-1}$ is:

Question

For the strong electrolytes NaOH, NaCl and $BaCl_{2}$ the molar conductances at infinite dilution are $248.1	imes10^{-4}$, $126.5	imes10^{-4}$ and $280	imes10^{-4}$ $sm^{2}mol^{-1}$ respectively, then $\lambda_{m}^{\circ}$ for $Ba(OH)_{2}$ in $sm^{2}mol^{-1}$ is:

Accepted Answer

To determine the molar conductance at infinite dilution, $\lambda_{m}^{\circ}$, for $Ba(OH)_{2}$, we can use the principle of additivity of molar conductances for strong electrolytes. According to this principle, the molar conductance at infinite dilution of a compound can be calculated by summing the molar conductances of its constituent ions.

Given:
- $\lambda_{m}^{\circ}$ for $NaOH = 248.1 \times 10^{-4} \, sm^{2}mol^{-1}$
- $\lambda_{m}^{\circ}$ for $NaCl = 126.5 \times 10^{-4} \, sm^{2}mol^{-1}$
- $\lambda_{m}^{\circ}$ for $BaCl_{2} = 280 \times 10^{-4} \, sm^{2}mol^{-1}$

We need to find $\lambda_{m}^{\circ}$ for $Ba(OH)_{2}$.

1. **Calculate the molar conductance of the hydroxide ion, $OH^{-}$:**

From $NaOH$, we have:
   $$
   \lambda_{m}^{\circ}(NaOH) = \lambda_{m}^{\circ}(Na^{+}) + \lambda_{m}^{\circ}(OH^{-})
   $$
   From $NaCl$, we have:
   $$
   \lambda_{m}^{\circ}(NaCl) = \lambda_{m}^{\circ}(Na^{+}) + \lambda_{m}^{\circ}(Cl^{-})
   $$

Subtracting these two equations gives:
   $$
   \lambda_{m}^{\circ}(NaOH) - \lambda_{m}^{\circ}(NaCl) = \lambda_{m}^{\circ}(OH^{-}) - \lambda_{m}^{\circ}(Cl^{-})
   $$

Substituting the given values:
   $$
   248.1 \times 10^{-4} - 126.5 \times 10^{-4} = \lambda_{m}^{\circ}(OH^{-}) - \lambda_{m}^{\circ}(Cl^{-})
   $$
   $$
   121.6 \times 10^{-4} = \lambda_{m}^{\circ}(OH^{-}) - \lambda_{m}^{\circ}(Cl^{-})
   $$

2. **Calculate the molar conductance of the barium ion, $Ba^{2+}$:**

From $BaCl_{2}$, we have:
   $$
   \lambda_{m}^{\circ}(BaCl_{2}) = \lambda_{m}^{\circ}(Ba^{2+}) + 2\lambda_{m}^{\circ}(Cl^{-})
   $$

Solving for $\lambda_{m}^{\circ}(Ba^{2+})$:
   $$
   \lambda_{m}^{\circ}(Ba^{2+}) = \lambda_{m}^{\circ}(BaCl_{2}) - 2\lambda_{m}^{\circ}(Cl^{-})
   $$

3. **Substitute $\lambda_{m}^{\circ}(Cl^{-})$ from the previous calculation:**

From the equation $\lambda_{m}^{\circ}(OH^{-}) - \lambda_{m}^{\circ}(Cl^{-}) = 121.6 \times 10^{-4}$, we can express $\lambda_{m}^{\circ}(Cl^{-})$ in terms of $\lambda_{m}^{\circ}(OH^{-})$.

4. **Calculate $\lambda_{m}^{\circ}(Ba(OH)_{2})$:**

$$
   \lambda_{m}^{\circ}(Ba(OH)_{2}) = \lambda_{m}^{\circ}(Ba^{2+}) + 2\lambda_{m}^{\circ}(OH^{-})
   $$

Substitute the values obtained from the above calculations to find $\lambda_{m}^{\circ}(Ba(OH)_{2})$.

After performing these calculations, we find that the correct value for $\lambda_{m}^{\circ}(Ba(OH)_{2})$ is indeed $523.2 \times 10^{-4} \, sm^{2}mol^{-1}$, which corresponds to Option 2.

**Explanation for Incorrect Options:**

- **Option 1: $401.6 \times 10^{-4}$**: This value is too low and does not account for the correct contributions of the ions involved.
- **Option 3: $776.2 \times 10^{-4}$**: This value is too high, suggesting an overestimation of the ionic contributions.
- **Option 4: $654.6 \times 10^{-4}$**: This value is also incorrect as it does not match the calculated sum of the ionic conductances.

Thus, the correct answer is Option 2: $523.2 \times 10^{-4} \, sm^{2}mol^{-1}$.