A stock solution containing

Mn2+

ions is prepared by dissolving 1.584 g of pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions are prepared by dilution. For solution A, 50.00 mL of stock solution is diluted to 1000.0 mL. For solution B, 10.00 mL of A is diluted to 250.0 mL. For solution C, 10.00 mL of B is diluted to 500.0 mL. Calculate the molar concentrations of the stock solution and solutions A, B, and C.

Question

Chemistry

Sneakers

23 July, 22:56

A stock solution containing

Mn2+

ions is prepared by dissolving 1.584 g of pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions are prepared by dilution. For solution A, 50.00 mL of stock solution is diluted to 1000.0 mL. For solution B, 10.00 mL of A is diluted to 250.0 mL. For solution C, 10.00 mL of B is diluted to 500.0 mL. Calculate the molar concentrations of the stock solution and solutions A, B, and C.

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Answers (1)

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Yaritza Reeves · Answer 1

The molar concentration of the stock solution is 0.0288 M.

The concentration of solution A is 1.440 * 10⁻³ M.

The concentration of solution B is 5.760 * 10⁻⁵ M.

The concentration of solution C is 1.152 * 10⁻⁶ M.

Explanation:

Hi there!

The molar mass of manganese is 54.938 g/mol, then, if we have 1.584 g in the stock solution, we will have (1.584 g · 1 mol / 54.938 g) 0.0288 mol manganese in 1 l.

Then, the molar concentration of the stock solution is 0.0288 M.

To calculate the concentration of each dilution, we have to take into account that the number of moles that is present in the volume taken from the concentrated solution will be the same as in the dilution. Then:

number of moles solution 1 (moles present in the taken volume) = number of moles of solution 2 (diluted solution).

number of moles 1 = number of moles 2

Concentration 1 · volume 1 = concentration 2 · volume 2

For solution A

Concentration solution 1 = 0.0288 M

Volume taken from solution 1 = 50.00 ml

Concentration solution 2 = unknown.

volume solution 2 = 1000.0 ml

Then:

0.0288 M · 50.00 ml = 1000.0 ml · Concentration 2

0.0288 M · 50.00 ml / 1000.0 ml = C₂

C₂ = 1.440 * 10⁻³ M

The concentration of the solution A is 1.440 * 10⁻³ M.

We proceed in the same way for the other dilutions:

Solution B:

C₁ · V₁ = C₂ · V₂

1.440 * 10⁻³ M · 10.00 ml = C₂ · 250.0 ml

1.440 * 10⁻³ M · 10.00 ml / 250.0 ml = C₂

C₂ = 5.760 * 10⁻⁵ M

The concentration of solution B is 5.760 * 10⁻⁵ M.

Solution C:

5.760 * 10⁻⁵ M · 10.00 ml = C₂ · 500.0 ml

5.760 * 10⁻⁵ M · 10.00 ml / 500.0 ml = C₂

C₂ = 1.152 * 10⁻⁶ M

The concentration of solution C is 1.152 * 10⁻⁶ M.