What is the molarity of sodium ions in a solution prepared by diluting 250. mL of 0.550 M Na2SO4 to 1.25 L? (A) 0.110 M (C) 0.220 M (B) 0.138 M (D) 0.275 M

We can simply adopt the mole concept in solving this dilution problem. During dilution, we intend to reduce the concentration of a particular solution by increasing its volume. Therefore, we have to find the new concentration of the diluted Na₂SO₄. From here, we can be able to find the molarity of the sodium ions in the diluted solution:

Using dilution equation: [tex]\frac{C_{1} }{V_{1} }[/tex] = [tex]\frac{C_{2} }{V_{2} }[/tex]

we can solve for the molarity of the diluted solution

C₁ = concentration of the original solution

V₁ = volume of the original solution

V₂ = final volume of the diluted solution

C₂ = final volume

We are solving of C₂:

Making C₂ the subject of the formula gives:

C₂ = [tex]\frac{C_{1}V_{1} }{V_{2} }[/tex]

Therefore:

C₂ = [tex]\frac{0.25 x 0.55 }{1.25 }[/tex] = 0.11M

The diluted concentration is now 0.11M

To find the molarity of Na⁺ in Na₂SO₄:

We write the ionic form of the compound in solution:

Na₂SO₄ → 2Na⁺ + SO₄²⁻

In this solution,

1 mole of Na₂SO₄ will produce 2 moles of Na⁺ ions

From 0.11M of Na₂SO₄, we would have (2 x 0.11M) = 0.22M of Na⁺

The molarity of sodium ions will be 0.22M