A stock solution containing Mn2+ ions was prepared by dissolving 1.166 g pure manganese metal in nitric acid?and diluting to a final volume of 1.000 L. The following solutions were then prepared by dilution. For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL. For solution B, 10.00 mL of solution A was diluted to 250.0 mL. For solution C, 10.00 mL of solution B was diluted to 500.0 mL. Calculate the molar concentrations of the stock solution and solutions A, B, and C.

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laurahai laurahai · Answer 1 · 2019-09-07T21:19:21+02:00

Answer:

Stock concentration: 0.02122 M

Solution A concentration: 1.061 x 10⁻³ M

Solution B concentration: 4.244 x 10⁻⁵ M

Solution C concentration: 8.488 x 10⁻⁷ M

Explanation:

The moles of manganese metal reacted is calculated by using the atomic weight (54.938 g/mol):

(1.166g) / (54.938 g/mol) = 0.02122 mol Mn = 0.02122 mol Mn²⁺

The manganese metal and Mn²⁺ ions are related 1:1. The concentration of the Mn²⁺ stock solution is then:

(0.02122 mol) / (1.000 L) = 0.02122 M Mn²⁺

The concentration of solution A can be found using the dilution equation:

C₂ = (C₁V₁)/V₂ = (0.02122M)(50.00mL) / (1000mL) = 1.061 x 10⁻³ M

Similarly for solution B, using the concentration solution A as C₁

C₂ = (C₁V₁)/V₂ = (1.061 x 10⁻³ M)(10.00mL) / (250.0mL) = 4.244 x 10⁻⁵ M

Similarly for solution C:

C₂ = (C₁V₁)/V₂ = (4.244 x 10⁻⁵ M)(10.00mL) / (500.0mL) = 8.488 x 10⁻⁷ M