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How many mL of 0.250 M MgCl2 must be added to 350.0 mL of 0.250 M NaCl to produce a solution in which the concentration of chloride ion is 0.300M Cl-?

  • chemistry -

    Remember M = moles/L (or mmoles/mL).

    So we start with how much Cl^-? That is 350 mL x 0.250M = 87.5 mmoles Cl^-. Let y = mL of the MgCl2 to be added.
    We have y mL of 0.500 M (in chloride) from MgCl2 to add to it. The total volume will be y mL + 350 mL. Now set that up as mmoles/mL and set it equal to 0.3M.
    0.5y mmoles Cl from MgCl2.
    350 x 0.250 M = 87.5 mmoles from NaCl.
    (0.5y + 87.5)/(y+350) = 0.3M
    Solve for y
    I estimated it as 87.5 mL of 0.250M MgCl2 (remember it's 0.500 M in Cl^-). You can always check that out. Post your work if you get stuck.

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