The concentration of Na⁺ ions in the final solution will be 0.33 M
Volume of Na₂SO₄ = 100 ml
Volume of NaCl = 200 ml
Molarity of Na₂SO₄ = 0.200 M
Molarity of NaCl = 0.300 M
Concentration of Na⁺ ions =?
We will calculate the number of moles of Na+ ions as
number of mole of Na⁺ in Na₂SO₄ = 0.100 L (2 × 0.200 mol Na₂SO₄ / L × 1 mol Na⁺ / 1 mol Na₂SO₄)
number of mole of Na⁺ in Na₂SO₄ = 0.04 mol Na⁺
number of mole of Na⁺ in NaCl = 0.200 L × 0.3 mol NaCl / L × mol Na⁺ / 1 mol NaCl
number of mole of Na⁺ in NaCl = 0.06 mol Na⁺
Now calculate the total volume
total volume = (0.100 + 0.200) L
total volume = 0.300 L
Finally, calculate the concentration
Concentration of Na⁺ = (0.04 + 0.06) mol Na+ / 0.300 L
Concentration of Na⁺ = 0.33 M
You can also learn about molarity from the following question:
https://brainly.com/question/16380318
#SPJ4
