The minimum volume of water required to dissolve 0.1 g lead(ii) chloride to get a saturated solution
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The minimum volume of water required to dissolve 0.1 g lead(ii) chloride will be 0.022 ltrs
Explanation:
we know that the solubility product Ksp of PbCl₂ = 1.7 x 10⁻⁵
Let S be the solubility of PbCl₂
concentration of reaction is given as:
PbCl₂ <-----------> Pb²⁺ + 2Cl⁻
[1-S] [S] [2s]
Hence
Ksp = [Pb²⁺][Cl⁻]² = S x [2S]² = 4S³
=> 4S³ = 1.7 x 10⁻⁵
=> S³ = 4.25 x 10⁻⁶
=> S = ∛4.25 x 10⁻⁶ = 1.62 x 10⁻²
Molar mass of PbCl₂ = 278
Therefore solubility of PbCl₂ in g/L = 1.62 x 10⁻² x 278 = 4.5
4.5 gms of PbCl₂ needs to be dissolved in 1 ltr
=> 0.1 gms of PbCl₂ will be dissolved in = 0.1/4.5 = 0.022 ltrs of water.
Hence The minimum volume of water required to dissolve 0.1 g lead(ii) chloride will be 0.022 ltrs
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