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TBR chem explanation

Optimist Prime

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5+ Year Member
Jun 23, 2015
  1. Medical Student (Accepted)
    What is the [Cl-] after mixing 50ml .1M KCL with 25ml .2M MGCl2?

    I can do it using an ice table but that just takes way too much time TBR's explanation on averaging the solutions is not explained at all. Can anyone help?


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    Jan 16, 2017
    1. MD/PhD Student
      The number of moles of Cl- will be conserved when mixing the solutions, so you need to find the total number of moles of Cl and then divide by the new volume (75 mL) after solutions are mixed. Following the units is useful in this situation as well.

      0.050 L * 0.1 mol/L * 1 (equivalent Cl in KCl) + 0.025 * 0.2 mol/L * 2 (equivalent Cl in MgCl2) = 0.015 mol Cl-

      0.015 mol Cl- / 0.075 L = 0.2 M Cl-

      shorthand version: as all volume here is in mL here, you don't really have to convert into liters. The order of terms is similar to the above.
      (50*0.1*1+25*0.2*2)/75 = 0.2 M Cl-

      This method works when both substances readily dissolve. If you were using something that didn't fully dissolve, such as AgCl, then you would have to do a ksp calculation
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      Aug 6, 2014
        Thanks! I was hoping the answer choices would allow for a "shortcut" to the answer, but that's not the case here. I would guess that TBR's reference to "averaging the solutions" simply means (total moles) / (total volume), as described by Asteri, which is probably the most straightforward way of solving this question.

        However, here's a different approach that's a little more conceptual and easier on the math. Maybe it'll be faster for you, maybe not.

        The first solution is 0.1 M KCl, which is effectively 0.1 M Cl-
        The second solution is 0.2 M MgCl2, which is effectively 0.4 M Cl-
        The molarity of Cl- of the mixed solution is the average of 0.1 and 0.4 M weighted by their respective volumes (50 and 25 mL). Notice the 2:1 ratio of these volumes; this is 2 parts 0.1 M and 1 part 0.4 M. This means that the answer will be closer to 0.1 M than it will be to 0.4 M (which unfortunately cancels out only 0.25 M, I was hoping for more!)

        If you simply calculate the average, [2(0.1)+(0.4)]/3 = 0.2
        If you're into fractions, the average is one third the way between 0.1 and 0.4 M, or 0.2 M
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