If 200.mL of 0.100 M Na₂SO₄ is added to 200.mL of 0.150 M NaCl,what is the concentration of Na⁺ ions in the final solution? Assume that the volumes are additive. A)0.05 M B)0.175 M C)0.125 M D)0.250 M

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The concentration of Na+ ions in the final solution is calculated by considering the total moles of Na+ ions from both Na2SO4 and NaCl and the total volume of the solution. Initially, there are 0.020 moles of Na+ from Na2SO4 (since 0.100 M * 0.200 L = 0.020 moles and each formula unit of Na2SO4 contributes 2 Na+ ions, thus 0.010 moles of Na2SO4 * 2 = 0.020 moles of Na+) and 0.030 moles of Na+ from NaCl (0.150 M * 0.200 L = 0.030 moles). The total moles of Na+ ions is 0.020 + 0.030 = 0.050 moles. The total volume of the solution is 400 mL or 0.400 L. The concentration of Na+ ions is then 0.050 moles / 0.400 L = 0.125 M.

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