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Calculating the molarity when mixing two different solutions

  1. Apr 6, 2016 #1
    1. The problem statement, all variables and given/known data
    You mix 45-mL of 0.1174M K2SO4 and 35-mL of 0.2504M HNO3

    2. Relevant equations
    none.

    3. The attempt at a solution

    So i'm supposed to find the number of moles of the K2SO4 and add it to the number of moles of 0.2504, then add the two volumes, and divide the total number of moles by the total volume.

    My question is, if i write a balanced equation, i get K2SO4 + 2HNO3 = 2KNO3 + H2SO4, does that mean for the number of moles of HNO3, before i add it to the number of moles of K2SO4 i have to multiply the number of moles of HNO3 by 2, because of the coefficient.

    I could also be doing this problem entirely wrong by assuming a reaction is occurring and even needing to balance. There is nothing in my book talking about reactions when mixing two different solutions.
     
  2. jcsd
  3. Apr 7, 2016 #2

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    That's the complete problem statement?
     
  4. Apr 7, 2016 #3
    In this case the problem said to find the molarity of each ion present in the solution formed. But i can already do that, and on my test tomorrow theres going to be a question that has a part a) and b) (professor told us) and its gonna ask for the ion concentration and the molarity of final solution.

    Now you got me thinking that it isn't possible to do what i'm doing when given two different solutions, and that you can only find the ion concentration when mixing two different solutions, not the molarity. Is this true? If thats true then my professor might have meant mixing two same solutions of different molarity.
     
  5. Apr 7, 2016 #4

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    That would be my take on it.
     
  6. Apr 7, 2016 #5
    yeah you're definitely right, i just realized how it doesn't even make sense to have one molarity for a mixture of solutions
     
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