Question

If 100. ml of 0.400 m Na₂SO₄ is added to 200. ml of 0.600 m NaCl, what is the concentration of Na ions in the final solution?

1) 1.00 m
2) 0.534 m
3) 1.40 m
4) 0.667 m

Answer 1

The concentration of Na ions in the final solution is 0.667 M.

Step-by-step explanation:

The concentration of Na ions in the final solution can be calculated by adding up the moles of Na ions from Na₂SO₄ and NaCl, and dividing by the final volume of the solution.

First, calculate the moles of Na ions in 100 ml of 0.400 M Na₂SO₄:

1. Concentration of Na in Na₂SO₄ = 0.400 M imes 2 = 0.800 M
2. Volume of Na₂SO₄ used = 100 ml = 0.100 L
3. Moles of Na in Na₂SO₄ = Concentration imes Volume = 0.800 M imes 0.100 L = 0.080 mol

Next, calculate the moles of Na ions in 200 ml of 0.600 M NaCl:

1. Concentration of Na in NaCl = 0.600 M imes 1 = 0.600 M
2. Volume of NaCl used = 200 ml = 0.200 L
3. Moles of Na in NaCl = Concentration imes Volume = 0.600 M imes 0.200 L = 0.120 mol

Finally, add up the moles of Na ions from Na₂SO₄ and NaCl and divide by the final volume:

1. Total moles of Na ions = Moles of Na in Na₂SO₄ + Moles of Na in NaCl = 0.080 mol + 0.120 mol = 0.200 mol
2. Final volume of the solution = Volume of Na₂SO₄ + Volume of NaCl = 0.100 L + 0.200 L = 0.300 L
3. Concentration of Na ions = Total moles / Final volume = 0.200 mol / 0.300 L = 0.667 M

Therefore, the concentration of Na ions in the final solution is 0.667 M.