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28 December, 22:34

A 25.00 mL solution of 0.150 M NaCl is combined with 10.00 mL of a 0.0750 M CaCl2 solution. Assuming a total volume of 35.00 mL, determine the concentration of chloride ion in the combined solution.

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  1. 29 December, 01:49
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    [Cl⁻] = 1,5x10⁻⁴M

    Explanation:

    First of all, let's determinate the mole of each salt.

    Molarity. volume = Mole

    Volume must be in L, cause molarity is mol/L

    NaCl → Na⁺ + Cl⁻

    Ratio is 1:1

    0.15 mol/L. 0.025L = 3.75x10⁻³ mole

    As ratio is 1:1, from 3.75x10⁻³ mole of salt, I have 3.75x10⁻³ mole of chloride

    CaCl₂ → Ca²⁺ + 2Cl⁻

    Ratio is 1:2 so, from 1 mol of salt I'll get the double of mole of chloride

    0.075 mol/L. 0.010 L = 7.5x10⁻⁴ mol

    7.5x10⁻⁴ mol. 2 = 1.5x10⁻³ mole

    Total mole of Cl⁻: 3.75x10⁻³ + 1.5x10⁻³ = 5.25x10⁻³

    This 5.25x10⁻³ mole are present in a total volume of 35 mL.

    Let's convert 35 mL in L → 0.035L (35/1000)

    Molarity is mol/L → 5.25x10⁻³ mol / 0.035L = 1,5x10⁻⁴M
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