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7 September, 03:10

Calculate the concentration of chloride ions when 100.0 mL of 0.233 M sodium chloride is mixed with 250.0 mL of 0.150 M calcium chloride.

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  1. 7 September, 03:43
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    The dissociation of both salts NaCl and CaCl₂ are as follows;

    NaCl - - > Na⁺ + Cl⁻

    CaCl₂ - - > Ca²⁺ + 2Cl⁻

    the molar ratio of NaCl to Cl⁻ is 1:1

    therefore number of NaCl moles is equal to number of Cl⁻ ions dissociated from NaCl

    then number of Cl⁻ ion moles - 0.233 mol/L x 0.1000 L = 0.0233 mol

    molar ratio of CaCl₂ to Cl⁻ ions is 1:2

    1 mol of CaCl₂ gives out 2 mol of Cl⁻ ions.

    number of CaCl₂ moles - 0.150 mol/L x 0.2500 L = 0.0375 mol

    then the number of Cl⁻ ion moles - 0.0375 x 2 = 0.0750

    total number of Cl⁻ ion moles = 0.0233 mol + 0.0750 mol = 0.0983 mol

    volume of solution - 100.0 + 250.0 = 350.0 mL

    concentration of Cl⁻ = 0.0983 mol / 0.3500 L = 0.281 M

    concentration of Cl⁻⁻ is 0.281 M
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