The conductivity of saturated solution of Ba₃ (PO₄)₂ is $1.2 X 10^{- 5} Ω^{- 1} c m^{- 1}$ . The limiting equivalent conductivities of BaCl₂, K₃PO₄ and KCl are 160, 140 and $100 Ω^{- 1} c m^{2} e q^{- 1}$ , respectively. The solubility product of Ba₃ (PO₄)₂ is

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10^-5

1.08 x 10^-23

1.08 x 10^-25

1.08 x 10^-27

answer is B.

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Detailed Solution

$\land_{cq}^{0} [{Be}_{3} {({PO}_{4})}_{2}] = \land_{cq}^{0} [{BeCℓ}_{2}] + \land_{cq}^{0} [K_{3} {PO}_{4}] + \land_{cq}^{0} [KCℓ]$
= 160 + 140 - 100 = 200ohm^-1cm²eq^-1
Now, $n_{eq} = \frac{k}{c} \Rightarrow 200 = \frac{1.2 \times 10^{- 5}}{c}$
$\Rightarrow$ c = 6 x 10^-8 eq. cm^-3 = 6 x 10^-5 N = 10^-5 M
Now, k_sp = 108s⁵ = 108(10^-5)⁵ = 1.08 x 10^-23