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Introductory Stoichiometry Question

  1. Feb 19, 2012 #1
    A solution of KOH is prepared by dissolving 2.00g of KOH in water to a final volume of 250mL of solution. What volume of this solution will neutralize 20.0mL of 0.115 mol/L sulfuric acid?


    2. Relevant equations
    C=n/v
    C1V1=C2V2
    n=m/M

    Attempt at solution

    First I balanced the equation:
    2KOH+H2SO4->k2S04+2H20

    Than I figured out the moles of H2S04:
    v=0.02L
    C=0.115mol/L
    n=c*V=2.3x10^-3 mols.

    Than I found the mols of KOH
    m=2g
    M=56.1g/mol
    n=m/M.
    n=0.0356 mols

    The problem is after this step I get confused as to what to do with the 250mL. Can someone give me an explanation please on how and why? (if you help me solve it)
     
  2. jcsd
  3. Feb 20, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    What is molarity of the KOH solution? What volume of this solution contains 0.0356 moles of KOH?
     
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