The conductivity of saturated solution of Ba3(PO4)2 is 1.2 × 10−5 Ω−1 cm−1. The limiting equivalent

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The conductivity of saturated solution of Ba3(PO4)2 is 1.2 × 10−5 Ω−1 cm−1. The limiting equivalent conductivities of BaCl2, K3PO4 and KCl are 160, 140 and 100 Ω−1 cm2 eq−1, respectively. The solubility product of Ba3(PO4)2 is

Accepted Answer

$\Lambda_{eq}^{o}\left\lbrack Be_{3}\left( Po_{4} \right)_{2} \right\rbrack = \Lambda_{eq}^{o}\left\lbrack BeCl_{2} \right\rbrack + \Lambda_{eq}^{o}\left\lbrack K_{3}Po_{4} \right\rbrack - \Lambda_{eq}^{o}\lbrack KCl\rbrack$

$= 160 + 140 - 100 = 200\text{ Oh}\text{m}^{- 1}\text{ c}\text{m}^{2}\text{ e}\text{q}^{- 1} $$${\text{Now, }n_{eq} = \frac{\kappa}{C} \Rightarrow 200 = \frac{1.2 \times 10^{- 5}}{C} }{\Rightarrow C = 6 \times 10^{- 8}\text{eq c}\text{m}^{- 3} }{= 6 \times 10^{- 5}\text{ N = 1}\text{0}^{- 5}\text{ M} }{\text{Now, }K_{sp} = 108S^{5} = 108 \times \left( 10^{- 5} \right)^{5} = 1.08 \times 10^{- 23}}$$

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