Question

Consider an aqueous solution of HCl and Ca(NO₃)₂ with the following concentrations: [HCl] = 0.30 M, [Ca(NO₃)₂] = 0.10 M.

(a) Calculate the conductivity of this solution at 25 ^oC.
(b) Calculate the resistance of a cylinder of this solution with the diameter of 2.0 cm and the length of 10 cm. Electrodes are located at the opposite ends of the cylinder.

Answer 1

a. Both substances are strong electrolytes (completely dissociated).

Val=equivalent

[H+] = 0.3 val/L

[Cl-] =0.3 val/L

[Ca2+] = 0.2 val/L

[NO₃^-] = 0.2 val/L

The equivalent ionic conductivities (from the table) have to be summed as λC to obtain the equivalent conductivity k of the solution:

k = 349.0x0.3 + 76.3x0.3 + 59.5x0.2 + 71.4x0.2 = 154 (mho.cm²/val).(val/1000 cm³)=

    = 0.154 mho.cm

b. The resistivity ρ is

ρ = 1/k = 1/0.154 mho.cm = 6.50 Ω/cm

The resistance R = ρ. l /A

R = 6.50 Ω/cm · 10 cm / (3.14x 1 cm²) = 21 Ω

Ionic Conductances, mho-cm2/val (I=0, 25^oC)

Cation	λ0+	Anion	λ0+
H+	349.8	OH-	198.0
Na+	50.1	HCO3-	44.5
K+	73.5	F-	55.4
Li+	38.7	Cl-	76.3
NH4+	73.4	Br-	78.4
Ca+2	59.5	CH3COO-	40.9
Mg+2	53.1	NO3-	71.4
		SO4-2	79.8