Find the molarity of Na+ ions of a diluted solution.

Question

Calculate the molarity of the Na⁺ ion in a solution resulting from the dilution of 25.1 mL of 5.12×10^-3 M Na₂S₂O₃ to a final volume of 264 mL.

Answer 1

Sure, here's the solution with the molarity of the Na⁺ ion added:

To calculate the molarity of the Na⁺ ion in the resulting solution, we first need to determine the number of moles of Na⁺ ions present in the initial solution and then use the dilution formula to find the final concentration.

Given:

- Initial volume (Vi) = 25.1 mL = 0.0251 L

- Initial molarity (Mi) of Na2S2O3 = 5.12×10⁻³ M

- Final volume (Vf) = 264 mL = 0.264 L

First, let's find the number of moles of Na2S2O3 initially:

Moles of Na2S2O3 = Mi × Vi

Moles of Na2S2O3 = 5.12 × 10⁻³ M × 0.0251 L

Moles of Na2S2O3 = 0.000128512 mol

Since there are 2 moles of Na⁺ ions for every 1 mole of Na2S2O3, the number of moles of Na⁺ ions is twice the number of moles of Na2S2O3:

Moles of Na⁺ = 2 × 0.000128512 mol

Moles of Na⁺ = 0.000257024 mol

Now, let's use the dilution formula to find the final molarity of Na⁺:

Mf = nf / Vf

Where:

- Mf = final molarity

- nf = final number of moles

- Vf = final volume

Since moles are conserved upon dilution:

nf = 0.000257024 mol

Vf = 0.264 L

Mf = nf / Vf

Mf ≈ 9.75 × 10⁻⁴ M

So, the molarity of the Na⁺ ion in the resulting solution is approximately 9.75 × 10⁻⁴ M.