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
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)