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    A student mixes 100 cm3
    of 0.200 mol dm–3 NaCl(aq) with 100 cm3
    of
    0.200 mol dm–3 Na2CO3(aq).
    What is the total concentration of Na+
    ions in the mixture formed?

    I know that the answer is 0.300 mol dm-3 but how do you work it out?
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    Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.

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    (Original post by 12aryana)
    A student mixes 100 cm3
    of 0.200 mol dm–3 NaCl(aq) with 100 cm3
    of
    0.200 mol dm–3 Na2CO3(aq).
    What is the total concentration of Na+
    ions in the mixture formed?

    I know that the answer is 0.300 mol dm-3 but how do you work it out?

    NaCl(aq) = Na+(aq) + Cl-(aq)

    Na2CO3(aq) = 2Na+(aq) + CO32-(aq)

    As you can see one mole of NaCl gives one mol of Na+ ions
    Also, one mole of Na2CO3 gives two mol of Na+ ions

    So from NaCl, using conc. x volume = moles

    0.2 x 0.1 = 0.02 moles

    From Na2CO3

    0.2 x 0.1 x 2 = 0.04 moles

    Therefore, total moles of ions

    0.04 + 0.02 = 0.06 moles

    This is in a total volume now of 0.2 dm-3 (2 x 100 cm3)

    Final concentration = 0.06/0.2 = 0.3 mol dm-3
 
 
 

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