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Math Help - calculating ion concentrations2

  1. #1
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    calculating ion concentrations2

    2.75g of MgO is added to 70mL of 2.40mol/L HNO3. Is the solution that results from the reaction acidic or basic? What is the concentration of the ion that is responsible for the character of the solution?
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  2. #2
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    The first step here is to convert the MgO to mol, so we divide by the RMM of MgO.
    <br />
\frac {2.75}{16.9994+24.3050} = 0.06658

    Then we divide by the volume in Litres to get molarity.
    <br />
\frac {.06658}{.070} = .9511

    In solution, the MgO splits up into Mg^{2+} and O^{2-} ions while the HNO_3 splits up into H^+ and NO_3^- ions.

    Since the O^{2-} ion readily accepts 2 protons, we need to double the molarity to get the number of protons accepted in each Litre of solution. This gives 1.902.

    Since the original concentration of H^+ was more than 1.902, the solution is acidic, so we are interested in [ H^+]

    This is simply 2.40-1.902 = .4977M = .50M to the correct number of significant figures.
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