Question

A stock solution containing Mn21 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 concentrations of the stock solution and solutions A, B, and C.

Answer 1

Answer:

The stock solution has a concentration of 0.021883 M

Solution 1 has a concentration of 0.0014415 M = 1.4415 * 10^-3 M

Solution 2 has a concentration of 0.00005766 M = 5.766 * 10^-5 M

Solution 3 has a concentration of 0.0000011532 M = 1.1532 * 10 ^-6 M

Explanation:

Step 1: Calculate moles of manganese

1.584 g / 54.94 g/mol

= 0.02883 moles Mn

Step 2: calculate molarity of stock solution

Molarity stock solution = moles / Litres

M = 0.02883 moles / 1.000 L

molarity Mn2+ = 0.02883 M

= 0.021883 M

Step 3: Calculate dilution 1

C1V1 = C2V2

C1 = initial stock conc = 0.02883 M

V1 = initial volume  = 50 mL = 50 ¨*10^-3

C2 = final conc  = TO BE DETERMINED

V2 = final volume  = 1000 mL

0.02883 * 50*10^-3 = C2 * 1L

C2 = 0.02883*50*10^-3 = 0.0014415 M

Step 4: Calculate dilution 2

C1V1 = C2V2

C1 = initial  conc = 0.0014415 M

V1 = initial volume  = 10 mL = 10 ¨*10^-3

C2 = final conc  = TO BE DETERMINED

V2 = final volume  = 250 mL = 250 *10^-3 L

C2 = (0.0014415 M*10 ¨*10^-3)/250 *10^-3 L = 0.00005766 M

Step 5: Calculate dilution 3

C1V1 = C2V2

C1 = initial  conc = 0.00005766 M

V1 = initial volume  = 10 mL = 10 ¨*10^-3

C2 = final conc  = TO BE DETERMINED

V2 = final volume  = 500 mL = 500 *10^-3 L

C2 = (0.00005766 M*10 ¨*10^-3)/500 *10^-3 L = 0.0000011532 M

The stock solution has a concentration of 0.021883 M

Solution 1 has a concentration of 0.0014415 M = 1.4415 * 10^-3 M

Solution 2 has a concentration of 0.00005766 M = 5.766 * 10^-5 M

Solution 3 has a concentration of 0.0000011532 M = 1.1532 * 10 ^-6 M