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Stoichiometry, concentration of solutions: Ba(OH)2 and H3PO4

  1. Mar 25, 2013 #1
    1. The problem statement, all variables and given/known data

    What volume of 0.0521 M Ba(OH)2 is required to react completely with 24.50 mL of 0.141 M H3PO4? Phosphoric acid is a triprotic acid.

    Answer: volume of 0.0521 M Ba(OH)2 required = 99.5 mL

    2. The attempt at a solution
    1. I begin by balancing the equation: 3Ba(OH)2 + 2H3PO4 → Ba3(PO4)2 + 6H2O

    2. I use M = Molarity = mol solute / L solution to find the mols of H3PO4... mol = 0.141 × 24.50 mL = 3.45 mol

    3. According to the mole ratio... 3 × 3.45 mol of Ba(OH)2 = 10.36 mol of Ba(OH)2 needed.

    4. M = mol/L ... mL = 10.36 mol / .0521 M = 199.0 mL of Ba(OH)2 needed, which is a wrong answer.

    These are the steps I am taking. I must be making a mistake since I know that the answer is 99.5 mL.
     
  2. jcsd
  3. Mar 25, 2013 #2
    Check this again.
     
  4. Mar 26, 2013 #3

    Borek

    User Avatar

    Staff: Mentor

    Note that your answer is exactly twice too large (which points in exactly the same direction Pranav-Arora alredy signaled).
     
  5. Mar 26, 2013 #4
    I see it now. The mole ratio is 3/2.

    Thank you both.
     
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