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# Chemistery

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1. 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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Just quoting in Puddles the Monkey so she can move the thread if needed
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3. (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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