The concentration of the sodium ion in final solution containing 100 ml of 0.200 m of Na₂SO₄ and 200 ml of 0.300 m of NaCl is 20.48 M.
The concentration of Na₂SO₄ is given as 0.200 M. So it is clear that 0.200 mol of Na₂SO₄ is present in 1 L of the solution.
One mole of the Na₂SO₄ contains 2 moles of the Na⁺ ions and one mole of SO₄²⁻ ion.
Therefore, the concentration of Na⁺ ion can be calculated as follows as:
Concentration of Na⁺ ions (M₁) = 2×0.200 M
= 0.4 M
Volume of the Na₂SO₄ (V₁) = 100 mL
Similarly, 0.300 M of NaCl is taken, which implies that the 0.300 moles of NaBr are present in the 1 L of solution.
One mole of the NaCl contains 1 mole of Na⁺ ions and one mole of Cl⁻ ion.
Therefore ,concentration of Na⁺ ion (M₂) = 0.300 M
Volume of the NaCl (V₂) = 200 mL
It is known that, M₁V₁ + M₂V₂ = MV
Where V is the total volume of solution, that is 100 mL+200 mL = 300 mL and M is the concentration of Na⁺ ion in final solution. Equating values of M₁, V₁, M₂, and V₂ in above equation is as follows:
0.2 M×100 mL +0.150 M×200 mL = M×300 mL
M=0.2 M × 100 mL + 0.300 M × 200 mL/ 300 mL
M= 20.48 M
M= 20.48 M
Hence, concentration of Na⁺ ion in the final solution is 20.48 M.
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