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Calculate the concentration (in M) of Cl- ions in solution C

  1. Dec 8, 2015 #1
    1. The problem statement, all variables and given/known data

    PROBLEM #1 You have 3.00 L of a 3.39 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 Msolution of AgNO3(aq) called solution B. You mix these solutions together, making solution C.

    Calculate the concentration (in M) of Cl- ions in solution C.


    PROBLEM #2 You have 3.00 L of a 3.00 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 Msolution of AgNO3(aq) called solution B. You mix these solutions together, making solution C.

    Calculate the concentration (in M) of Na+ ions in solution C.

    2. Relevant equations
    nope

    3. The attempt at a solution
    For problem #2, i figured it out easily. I just did:
    3.00M x 3.00L = 9moles.
    9moles/(2.00L + 3.00L) = 1.80M

    For problem #1, i found the answer but i dont understand why. Heres what its supposed to be like:

    3.39M x 3.00L = 10.17moles Nacl.
    2.00M x 2.00L = 4moles AgNO3 <------DON'T UNDERSTAND WHY IM FINDING BOTH FOR THIS ONE AND NOT THE OTHER

    now you do 10.17 moles NaCl - 4moles AgNO3 = 6.17Moles, (no idea why im subtracting for this and not the other).

    Now divide 6.17 by the total like the other problem. 6.17/(2.00L + 3.00L) = 1.234M...

    - i dont think its because one is a cation and one is an anion
    - my guess is that its because AgCl is a solid and NaNO3 is aqueous
     
    Last edited: Dec 8, 2015
  2. jcsd
  3. Dec 8, 2015 #2

    DrClaude

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    Staff: Mentor

    Yes. Write down the balanced equation for what happens when you mix NaCl and AgNO3, with proper labels [(aq), etc.], and see what is in solution.
     
  4. Dec 8, 2015 #3

    Borek

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    Staff: Mentor

    For the record: that's not a correct answer.
     
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