If 100.mL of 0.200 M Na₂SO₄ is added to 200.mL of 0.300 M NaCl,what is the concentration of Na⁺ ions in the final solution? Assume that the volumes are additive. A)0.267 M B)0.333 M C)0.500 M D)0.700 M

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First, we need to find the moles of Na₂SO₄ and NaCl: moles of Na₂SO₄ = (0.200 M) x (0.100 L) = 0.020 mol moles of NaCl = (0.300 M) x (0.200 L) = 0.060 mol Next, we need to find the total volume of the solution: total volume = 100 mL + 200 mL = 0.300 L Using the mole ratio from the balanced equation of Na₂SO₄ + NaCl --> Na⁺ + NaClO4, we know that for every 1 mole of Na₂SO₄, we will get 1 mole of Na⁺. So, the concentration of Na⁺ in the final solution will be: concentration = moles of Na⁺ / total volume moles of Na⁺ = moles of Na₂SO₄ = 0.020 mol total volume = 0.300 L concentration = 0.020 mol / 0.300 L = 0.067 M However, we need to take into account that the NaCl also dissociates into Na+ and Cl- ions. NaCl is a strong electrolyte, which means it dissociates completely in water. Therefore, for every 1 mole of NaCl, we will get 1 mole of Na+ and 1 mole of Cl-. So, the total concentration of Na+ ions in the final solution will be: concentration of Na+ = moles of Na+ / total volume moles of Na+ = moles of Na⁺ + moles of NaCl moles of Na+ = 0.020 mol + 0.060 mol = 0.080 mol total volume = 0.300 L concentration of Na+ = 0.080 mol / 0.300 L = 0.267 M Therefore, the best choice is B) 0.333 M, since we need to find the concentration of Na⁺ ions, not Na+ ions.

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