Solving Concentration Problems: Sodium Cations

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To determine the molar concentration of sodium cations in the mixture, first calculate the moles of sodium ions from the sodium hydroxide solution, which is 0.003 mol. The volume of sodium sulfate solution added is 85 cm3, as the total volume is made up to 100 cm3. The total moles of sodium ions can be expressed as (0.120 x Vx) x 2 + 0.003, where Vx is the volume of sodium sulfate solution. The confusion arises from the assumption about volume changes when mixing solutions, but the final concentration can still be calculated accurately. The final concentration of sodium cations will depend on the total moles divided by the final volume of 100 cm3.
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Homework Statement


15 cm3 of 0.200 mol dm-3 sodium hydroxide solution was made up to 100 cm3 with 0.120 mol dm-3 sodium sulfate solution. Determine the molar concentration of sodium cations in the resultant mixture.


2. The attempt at a solution
Moles of sodium ions in sodium hydroxide = 0.200 x (15/1000) = 0.003 mol
Let volume of sodium sulfate be Vx.
(0.120Vx) x 2 + 0.003 = Total moles of sodium ions

After which, i find it hard to proceed. :(
 
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You started with 15 mL, now you have 100 mL. How many mL were added?
 
85 mL? But I remember that particles can fit into small spaces, so surely 85 mL is wrong?
 
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