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# AS chemistry help please Watch

1. I've been trying to do one of the multiple choice questions from the ocr specimen paper (new spec), but have no clue how to go about it:
A student mixes 100 cm^3 of 0.200 mol dm^-3 NaCl (aq) with 100 cm^3 of 0.200 mol dm^-3 Na2CO3 (aq). What is the total concentration of Na+ ions in the mixture formed?
The answer ends up being 0.300 mol dm^3.
2. n(NaCl) = c x V(decimetres3)
= 0.200 x (100/1000)
= 0.02 mol
There is one Na+ ion per unit, so n(Na+) = n(NaCl) = 0.02 mol

n(Na2CO3) = 0.200 x (100/1000)
= 0.02 mol
There are two Na+ ions per unit, so n(Na+) = 2 x n(Na2CO3) = 0.04 mol

Total number of Na+ ions = 0.02 + 0.04
= 0.06 mol

Therefore concentration of mixture = n/V
There is now 200cm3 on total, which is 0.2 dm3, so:
0.06 mol / 0.2 dm3 = 0.300 mol dm-3
3. Thank you!

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Updated: April 9, 2016
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