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28 July, 14:10

The ksp of lead (ii) hydroxide, pb (oh) 2, is 1.43 * 10-20. calculate the solubility of this compound in g/l.

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  1. 28 July, 16:31
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    When Pb (OH) ₂ dissolves it dissociates as follows;

    Pb (OH) ₂ - - - > Pb²⁺ + 2OH⁻

    molar solubility is the number of moles of compound that can be dissolved in 1 L of solution.

    if molar solubility of Pb (OH) ₂ is x then molar solubility of Pb²⁺ is x and OH⁻ is 2x

    the formula for solubility product constant is as follows;

    ksp = [Pb²⁺][OH⁻]²

    ksp = (x) (2x) ²

    ksp = 4x³

    ksp = 1.43 x 10⁻²⁰

    4x³ = 1.43 x 10⁻²⁰

    x = 1.53 x 10⁻⁷ M

    molar solubility of Pb (OH) ₂ is 1.53 x 10⁻⁷ M

    molar mass is 241.2 g/mol

    solubility of Pb (OH) ₂ is 1.53 x 10⁻⁷ M x 241.2 g/mol = 3.69 x 10⁻⁵ g/L
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