Chemistry, asked by pavithkalyan5170, 1 year ago

The minimum volume of water required to dissolve 0.1 g lead(ii) chloride to get a saturated solution

Answers

Answered by shubhamjoshi033
2

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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