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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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