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A 0.110-L sample of an unknown HNO3 solution required 52.1 mL of 0.150 M Ba(OH)2 for complete neutralization. What was the concentration of the HNO3 solution?

  • Chemistry -

    52.1 mL of .150M solution contains
    .0521*.150 = .00782 moles of Ba(OH)2

    Since the reaction is

    2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2H2O

    we need .0156 moles of HNO3

    .0156moles/.110L = 0.142moles/L = .142M

  • Chemistry -

    2HNO3 + Ba(OH)2 ==> 2H2O + Ba(NO3)2
    mols Ba(OH)2 = M x L = ?
    Use the equation coefficients to convert mols Ba(OH)2 to mols HNO3. Note: that's mols HNO3 = 2*mols Ba(OH)2.
    Then M HNO3 = mols HNO3/L HNO3. You know L and mols, solve for M.

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