Excess sodium sulfate is added to exactly 50 mL of a solution containing an
unknown amount of silver(I) cation. Determine the concentration of the
silver ion in this solution if 3.54 mg of a solid is isolated.
Convert 3.54 mg AgCl to moles AgCl, then to moles Ag, then M Ag = moles/L soln.
.0533
To determine the concentration of the silver ion in the solution, we first need to understand the chemistry behind the reaction.
The reaction between sodium sulfate and silver(I) cation can be written as:
2Ag⁺ + SO₄²⁻ → Ag₂SO₄
From the balanced equation, we can see that 2 moles of silver(I) cation react with 1 mole of sodium sulfate to form 1 mole of silver sulfate.
Given that 3.54 mg of silver sulfate is isolated, we can now calculate the moles of silver sulfate:
First, convert the mass of silver sulfate to moles by using its molar mass. The molar mass of Ag₂SO₄ is:
2(Ag) + 1(S) + 4(O) = 2(107.87 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 311.87 g/mol
Now, calculate the moles of Ag₂SO₄:
moles = mass / molar mass = 3.54 mg / 311.87 g/mol = 0.01136 mol
Since 1 mole of Ag₂SO₄ is formed from 2 moles of Ag⁺ ions, we can calculate the moles of Ag⁺ ions:
moles of Ag⁺ = 0.01136 mol * (2 mol Ag⁺ / 1 mol Ag₂SO₄) = 0.02272 mol
Now that we know the moles of Ag⁺ ions, we can calculate their concentration in the solution.
The volume of the solution is given as 50 mL. To convert this to liters:
volume = 50 mL = 50 mL * (1 L / 1000 mL) = 0.050 L
Finally, calculate the concentration of Ag⁺ ions in the solution:
concentration = moles of Ag⁺ ions / volume of solution
= 0.02272 mol / 0.050 L
= 0.4544 mol/L
Therefore, the concentration of the silver ion in the solution is 0.4544 mol/L.