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20 May, 16:35

Equal volumes of 0.140 M AgNO3 and 0.200 M ZnCl2 solution are mixed. Calculate the equilibrium concentrations of Ag + and Zn2+.

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  1. 20 May, 16:57
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    The concentration of Zn²⁺ = 0.065M

    Explanation:

    2AgNO₃ + ZnCl₂ → 2AgCl + Zn (NO₃) ₂

    According to the balanced equation, 2 moles of AgNO₃ reacts with 1 mole of ZnCl₂ to form 2 moles of AgCl and 1 mole of Zn (NO₃) ₂.

    Molarity of AgNO₃ is 0.14M

    Molarity of ZnCl₂ is 0.2M

    0.14 M of AgNO₃ would react completely with 0.14/2) = 0.07 M of ZnCl₂, but the ZnCl₂ is more concentrated than that, so ZnCl₂ is in excess and AgNO₃ is the limiting reactant.

    Neglecting the Ksp of AgCl:

    All of the Ag is in the precipitate so the concentration of Ag⁺ is zero.

    (0.2 M ZnCl2 originally) - (0.07 M ZnCl2 reacted) = 0.13 M ZnCl₂ = 0.13 M Zn²⁺ of the original ZnCl₂ solution.

    But the original solution has been diluted by the addition of an equal volume of AgNO₃, so the concentration in the final solution is one-half of the original, so:

    0.13 M Zn²⁺ / 2

    = 0.065 M Zn²⁺

    AgCl (s) is also produced as a precipitate so the ions comprising AgCl are removed from solution.
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