Discussion
At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below:
| Salt | Molar conductivity (S cm2 mol−1) at concentration (M) | |
| 0.01 | 0.04 | |
| NaNO3 | 111 | 101 |
| NaCl | 117 | 107 |
| AgNO3 | 125 | 116 |
The conductivity of a saturated aqueous solution of AgCl is 1.40×10−6 S cm−1 at 300 K. If the solubility of AgCl in water at 300 K is X mol L−1, then log10(X−1) is
(Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)
Answer: C
The solubility is \(10^{-5}\) mol L−1, so \(\log_{10}(X^{-1})\) is 5.
For strong electrolytes Kohlrausch's law gives \(\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{c}\). With \(\sqrt{0.01} = 0.1\) and \(\sqrt{0.04} = 0.2\), each pair of values gives \(A\) and the limiting value:
\[ \begin{aligned} \mathrm{NaNO_{3}}:&\ A = 100,\ \Lambda^{\circ} = 121 \\ \mathrm{NaCl}:&\ A = 100,\ \Lambda^{\circ} = 127 \\ \mathrm{AgNO_{3}}:&\ A = 90,\ \Lambda^{\circ} = 134 \end{aligned} \]
By Kohlrausch's law of independent migration of ions, the nitrate and sodium contributions cancel:
\[ \begin{aligned} \Lambda^{\circ}_{\mathrm{AgCl}} &= \Lambda^{\circ}_{\mathrm{AgNO_3}} + \Lambda^{\circ}_{\mathrm{NaCl}} \\ &\quad - \Lambda^{\circ}_{\mathrm{NaNO_3}} \\ &= 134 + 127 - 121 = 140 \end{aligned} \]
in S cm2 mol−1.
\[ \begin{aligned} X &= \frac{1000\,\kappa}{\Lambda^{\circ}} = \frac{1000 \times 1.40 \times 10^{-6}}{140} \\ &= 1.0 \times 10^{-5}\ \text{mol L}^{-1} \end{aligned} \]
So \(\log_{10}(X^{-1}) = 5\). The factor 1000 converts cm3 to litres; misplacing it by a power of ten gives 4 or 6, and 3 is further off still.
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