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The solubility product of PbI 2 is 8 . 0 × 10 - 9 . The solubility of lead iodide in 0 . 1 molar solution of lead nitrate is x × 10 - 6 mol / L . The value of x is _________ (Rounded off to the nearest integer) [Given 2 = 1 . 41 ]

Correct answer

0

Step-by-step solution

Given: K sp PbI 2 = 8 × 10 - 9 To calculate : solubility of PbI 2 in 0 . 1 M solution of Pb NO 2 2 I   Pb NO 3 2 → Pb ( aq ) + 2 + 2 NO 3 - aq 0 . 1   M - - -                   0 . 1   M                   0 . 2   M II   PbI 2 s ⇌ Pb + 2 aq + 2 I - aq                                       &#16

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