Question

Two beakers are placed in a small closed container at 25 °C. One contains 256 mL of a 0.279 M aqueous solution of Na₂S₂O₃; the second contains 444 mL of a 0.120 M aqueous solution of Na₂S₂O₃. Small amounts of water evaporate from both solutions. As time passes, the volume of solution in the second beaker gradually _________increasesdecreases and that in the first gradually _________increasesdecreases. If we wait long enough, what will the final volumes and concentrations be?

	First beaker		Second beaker
Initial	256 mL	0.279 M	444 mL	0.120 M
Final	mL	M	mL	M

Answer 1

As time passes, the volume of solution in the second beaker gradually decreases and that in the first gradually increases.

	First beaker		Second beaker
Initial	256 mL	0.279 M	444 mL	0.120 M
Final	401 mL	0.178 M	299 mL	0.178 M

Explanation

The water from second beaker will evaporate and condense in the first beaker so that concentrations become equal.

For first beaker, concentration of Na₂S₂O₃ = 0.279 M

volume of solution = 256 mL

moles of Na₂S₂O₃ = (concentration of Na₂S₂O₃) * (volume of solution)

moles of Na₂S₂O₃ = (0.279 M) * (256 mL)

moles of Na₂S₂O₃ = 71.424 mmol

Similarly, for second beaker, moles of Na₂S₂O₃ = 53.28 mmol

Suppose volume decrease for second beaker = x

concentration of Na₂S₂O₃ in first beaker = concentration of Na₂S₂O₃ in second beaker

(moles of Na₂S₂O₃ in first beaker) / (volume of solution in first beaker) = (moles of Na₂S₂O₃ in second beaker) / (volume of solution in second beaker)

(71.424 mmol) / (256 mL + x) = (53.28 mmol) / (444 mL - x)

Solving for x, x = 145 mL

concentration of Na₂S₂O₃ in first beaker = (moles of Na₂S₂O₃ in first beaker) / (volume of solution in first beaker)

concentration of Na₂S₂O₃ in first beaker = (71.424 mmol) / (256 mL + 145 mL)

concentration of Na₂S₂O₃ in first beaker = 0.178 M