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10 May, 18:24

A solution is prepared by mixing 150.0 ml of 1.00*10^-2 m mg (no3) 2 and 250.0 ml of 1.00*10^-1m naf calculate the concentrations of mg2 + and f at equilibrium with solid mgf2 ksp = 6.4*10^-9

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  1. 10 May, 20:53
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    First, we have to get moles of Mg2 + = molarity * volume

    = 1 x 10^-2 * 0.15 L

    = 0.0015 mol

    initial [Mg2+] = 0.0015 / 0.4 L (total volume)

    = 0.00375 M

    then, moles of F - = molarity * volume

    = 1 x 10^-1 * 0.25

    = 0.025 Mol

    initial [F-] = 0.025 / 0.4 = 0.0625 M

    by using ICE table:

    MgF2 (s) → Mg2 + 2F-

    initial 0.00375 0.0625

    change + X + 2 X

    Equ 0.00375+X 0.0625+2X

    when Ksp = [Mg2+][F-]^2

    by substitution:

    6.4 x 10^-9 = (0.00375+X) (0.0625 + 2X) ^2 by solving for X

    ∴X = 0.00024 M

    ∴[Mg2+] = 0.00375 + X

    = 0.00375 + 0.00024

    = 0.004 M

    and [F-] = 0.0625 + 2X

    = 0.0625 + 2 * 0.00024

    = 0.06 M
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