Problem
A conductivity cell with cell constant \(K_{\text{cell}}=1.10\,\mathrm{cm^{-1}}\) is filled with a \(0.0250\,\mathrm{M}\) calcium acetate solution at \(25^\circ\mathrm{C}\). The measured resistance is \(R=185\,\Omega\). Determine the conductivity \(\kappa\) and the molar conductivity \(\Lambda_m\).
Key idea: Calcium acetate is an ionic compound; in water it dissociates into mobile ions that carry charge, producing measurable conductivity.
\[ \mathrm{Ca(C_2H_3O_2)_2(aq)\rightarrow Ca^{2+}(aq)+2\,C_2H_3O_2^{-}(aq)} \]
Solution
1) Convert resistance to conductance
Conductance \(G\) is the reciprocal of resistance:
\[ G=\frac{1}{R}=\frac{1}{185\,\Omega}=5.41\times 10^{-3}\,\mathrm{S}. \]
2) Compute the conductivity (specific conductivity) \(\kappa\)
Conductivity is related to conductance by the cell constant:
\[ \kappa=K_{\text{cell}}\cdot G. \]
Using \(K_{\text{cell}}=1.10\,\mathrm{cm^{-1}}\):
\[ \kappa=(1.10\,\mathrm{cm^{-1}})\cdot(5.41\times 10^{-3}\,\mathrm{S}) =5.95\times 10^{-3}\,\mathrm{S\,cm^{-1}}. \]
Convert \(\mathrm{S\,cm^{-1}}\) to \(\mathrm{S\,m^{-1}}\) using \(1\,\mathrm{S\,cm^{-1}}=100\,\mathrm{S\,m^{-1}}\):
\[ \kappa=(5.95\times 10^{-3})(100)=0.595\,\mathrm{S\,m^{-1}}. \]
3) Compute the molar conductivity \(\Lambda_m\)
Molar conductivity is conductivity divided by molar concentration expressed in \(\mathrm{mol\,m^{-3}}\). Convert \(0.0250\,\mathrm{M}\) to \(\mathrm{mol\,m^{-3}}\):
\[ c=0.0250\,\mathrm{mol\,L^{-1}}\times 1000\,\mathrm{L\,m^{-3}}=25.0\,\mathrm{mol\,m^{-3}}. \]
Then
\[ \Lambda_m=\frac{\kappa}{c}=\frac{0.595\,\mathrm{S\,m^{-1}}}{25.0\,\mathrm{mol\,m^{-3}}} =2.38\times 10^{-2}\,\mathrm{S\,m^2\,mol^{-1}}. \]
A common chemistry unit is \(\mathrm{S\,cm^2\,mol^{-1}}\). Using \(1\,\mathrm{S\,m^2}=10^{4}\,\mathrm{S\,cm^2}\):
\[ \Lambda_m=(2.38\times 10^{-2})(10^{4})=2.38\times 10^{2}\,\mathrm{S\,cm^2\,mol^{-1}} \approx 238\,\mathrm{S\,cm^2\,mol^{-1}}. \]
4) Interpretation for calcium acetate conductivity
- Calcium acetate dissociates into \(\mathrm{Ca^{2+}}\) and \(2\,\mathrm{C_2H_3O_2^-}\), increasing the number of charge carriers per formula unit.
- For an ideal strong electrolyte assumption, the ion concentrations are \([\mathrm{Ca^{2+}}]=0.0250\,\mathrm{M}\) and \([\mathrm{C_2H_3O_2^-}]=0.0500\,\mathrm{M}\); higher ionic concentration generally increases \(\kappa\).
- Molar conductivity \(\Lambda_m\) is useful for comparing solutions at different concentrations; it typically decreases as concentration increases due to interionic interactions, even when dissociation is essentially complete.
| Quantity | Result | Formula used |
|---|---|---|
| Conductance \(G\) | \(5.41\times 10^{-3}\,\mathrm{S}\) | \(G=1/R\) |
| Conductivity \(\kappa\) | \(5.95\times 10^{-3}\,\mathrm{S\,cm^{-1}}=0.595\,\mathrm{S\,m^{-1}}\) | \(\kappa=K_{\text{cell}}G\) |
| Concentration \(c\) | \(25.0\,\mathrm{mol\,m^{-3}}\) | \(0.0250\,\mathrm{M}\times 1000\) |
| Molar conductivity \(\Lambda_m\) | \(2.38\times 10^{-2}\,\mathrm{S\,m^2\,mol^{-1}}\approx 238\,\mathrm{S\,cm^2\,mol^{-1}}\) | \(\Lambda_m=\kappa/c\) |
Visualization
The diagram shows a conductivity cell: the measured resistance depends on the geometry (cell constant) and on how well ions in the calcium acetate solution carry charge between the electrodes.
\[ G=\frac{1}{R},\qquad \kappa=K_{\text{cell}}\cdot G,\qquad \Lambda_m=\frac{\kappa}{c}. \]
Final results
\[ \kappa=5.95\times 10^{-3}\,\mathrm{S\,cm^{-1}}=0.595\,\mathrm{S\,m^{-1}}, \qquad \Lambda_m=2.38\times 10^{-2}\,\mathrm{S\,m^2\,mol^{-1}}\approx 238\,\mathrm{S\,cm^2\,mol^{-1}}. \]