Lt asked in Science & MathematicsChemistry · 9 years ago

Chemistry kinetics question?

How do you do this?

A NaCl solution (100 mL) is mixed with a AgNO3 solution (100 mL) to initiate a reaction. The solution of NaCl had an original concentration of 1.5 mol L-1. Immediately after mixing the two solutions, what was the concentration of NaCl

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  • 9 years ago
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    You mean before the AgNO3 began to react with NaCl to produce AgCl which is really insoluble?

    The reaction is very fast. So I will answer the question just as it is written.

    What will happen when the volume is doubled is that the concentration is halved. 1.5/2 = 0.75

    You can show this by using the dilution equation.

    V1*C1 = V2*C2

    V1 = 100 mL = 0.1 L

    V2 = 100 mL + 100 mL = 200 mL = 0.2 L

    C1 = 1.5 mol/L

    C2 = ???

    C2 * 0.2 = 1.5*0.1

    C2 = 0.75 mol/L

    =======

    If you mean after the AgCl was formed then you need more information. You need to know the concentation of AgNO3.

  • Anonymous
    9 years ago

    You make two approximations to get an answer

    1. You assume that 100 ml + 100 ml = 200 ml. The more dilute the two solutions, the more true that would be. But it is a good assumption, anyway.

    2. You assume that you can mix the two to a homogeneous solution "instantly" when compared to the precipitation rate of AgCl. This will be an approximation and not all that accurate, but any computations for the actual NaCl concentration will be wrong since until the solution is fully mixed you can't meaningfully speak about "the concentration" (since concentration will not be unique, rather on a distribution between 0 and 1.5 M).

    So, just divide by two.

  • 9 years ago

    it was simply diluted 2 fold

    if I have 100 ml of X and I pour in 100 ml of another solution

    I now have 200 ml total, with 100 ml of X

    so my concentration is 50% of my original (in this problem it would be .75)

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