ElectrochemistryHard

Question

Calculate the ionic product of water at 25°C from the following data:

Conductivity of water = 5.5 × 10−6 mho m−1

$\lambda_{H^{+}}^{o}$= 0.035 mho m2 mol−1 $\lambda_{OH^{-}}^{o}$= 0.020 mho m2 mol−1

Options

A.2 × 10−14 M2
B.1 × 10−7 M
C.1 × 10−8 M2
D.1 × 10−14 M2

Solution

$\Lambda_{m\left( H_{2}O \right)}^{o} = \lambda_{m\left( H^{+} \right)}^{o} + \lambda_{m\left( OH^{-} \right)}^{o} = 0.035 + 0.020 = 0.055\text{ mho }\text{m}^{2}\text{ mo}\text{l}^{- 1}$

Now, $\Lambda_{m} = \frac{k}{C} \Rightarrow 0.055 = \frac{5.5 \times 10^{- 6}}{C}$

$\therefore C = 10^{- 4}\text{ mol }\text{m}^{- 3} = 10^{- 7}\text{ M} \Rightarrow K_{w} = C^{2} = 10^{- 14}\text{ }\text{M}^{2}$

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