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The conductivity of saturated solution of `Ba_(3) (PO_(4))_(2)` is `1.2 xx 10^(-5) Omega^(-1) cm^(-1)`. The limiting equivalent conductivities of `BaaCl_(2), K_(3)PO_(4)` and KCl are `160, 140` and `100 Omega^(-1) cm^(2)eq^(-1)`, respectively. The solubility product of `Ba_(3)(PO_(4))_(2)` isA. `10^(-5)`B. `1.08 xx 10^(-23)`C. `1.08 xx 10^(-25)`D. `1.08 xx 10^(-27)` |
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Answer» Correct Answer - B `Lambda_(M)^(@) = Lambda_(n)^(@) xx n - ` Factor `Lambda_(M)^(@) (BaCl_(2)) = 160 xx 2 = 320` `Lambda_(M)^(@) (K_(2)PO_(4)) = 140 xx 3 420` `Lambda_(M)^(@) (KCl) = 100 xx 1 = 100` `Lambda_(M_([Ba_(2)(PO_(4))_(2)])^(@) = 3 Lambda_(M(BaCl_(2)))^(@) + 2 Lambda_(M(K_(3)PO_(4)))^(@) - 6 Lambda_(M(HCl))^(@)` `3 xx 320 + 2 xx 420 - 6 xx 100` `= 1200` `Lambda_(M)^(@) = (K xx 1000)/(s)` `s = (1.2 xx 10^(-5) xx 1000)/(1200)` `1 xx 10^(-5)` `K_(sp[Ba(PO_(4))_(2)]) = 108 s^(5)` `108 xx (10^(-5))^(5)` `= 1.08 xx 10^(-23)` |
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