Final answer:
After mixing 200cm³ of 0.2M Na₂SO₄ with 200cm³ of water, the concentration of Na+ is 0.2M and the concentration of SO4²− is 0.1M.
Step-by-step explanation:
To calculate the concentration of Na+ and SO42− ions after dilution, we first need to determine the number of moles of Na2SO4 in the original solution. Since the molarity is 0.2 M and the volume is 200 cm3, or 0.2 dm3, we have:
Number of moles of Na2SO4 = Molarity × Volume = 0.2 mol/dm3 × 0.2 dm3 = 0.04 moles
Each mole of Na2SO4 provides 2 moles of Na+ and 1 mole of SO42−. So, we have 0.08 moles of Na+ and 0.04 moles of SO42−. After dilution, the total volume of the solution is 200 cm3 + 200 cm3 = 400 cm3, or 0.4 dm3. The new concentrations are:
Concentration of Na+ = Moles of Na+ / Total Volume = 0.08 moles / 0.4 dm3 = 0.2 M
Concentration of SO42− = Moles of SO42− / Total Volume = 0.04 moles / 0.4 dm3 = 0.1 M