Question

A pipet is used to transfer 5.00 mL of a 1.25 M stock solution in flask "S" to a 25.00 mL volumetric flask "B," which is then diluted with DI H2O to the calibration mark. The solution is thoroughly mixed. Next, 2.00 mL of the solution in volumetric flask "A" is transferred by pipet to 50.00 mL volumetric flask "B" and then diluted with DI H2O to the calibration mark. Calculate the molarity of the solution in volumetric flask "B". How do I solve this?

Answer 1

Answer: the molarity of the solution in volumetric flask "B' is 0.0100 M

Explanation:

Given that;

the Molarity of stock solution M₁ = 1.25M

The molarity os solution in volumetric flask A (M₂) = M₂

Volume of stock solution pipet out (V₁) = 5.00mL

Volume of solution in volumetric flask A V₂ = 25.00mL

using the dilution formula

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

WE SUBSTITUTE

M₂ =  ( 1.25 × 5.00 ) / 25.00 mL

M₂ = 0.25 M

Now volume of solution pipet out from volumetric flask A V₂ = 2.00 mL

Molarity of solution in volumetric flask B (M₃) = M₃

Volume of solution in volumetric flask B V₃ = 50.00m L

Using dilution formula again

M₂V₂ = M₃V₃

M₃ = M₂V₂ / V₃

WE SUBSTITUTE

M₃ = ( 0.25 × 2.0) / 50.0

M₃ = 0.0100 M

Therefore the molarity of the solution in volumetric flask "B' is 0.0100 M

Answer 2

The concentration of the final solution is 0.01 M.

This is a problem of serial dilution. We have to first obtain the concentration of  the solution in the new flask.

C1V1 = C2 V2

C1 = concentration of stock solution = 1.25 M

V1 = volume of stock solution =  5.00 mL

C2 = concentration of solution in the new flask = ?

V2 = volume of solution in flask B in the new flask = 25.00 mL

C2 = C1V1 /V2

C2 = 1.25 M ×  5.00 mL/ 25.00 mL

C2 = 0.25 M

Again we need to find the concentration when this solution is further diluted;

C1 = 0.25 M

V1 = 2.00 mL

C2 = ?

V2 = 50.00 mL

C2 = C1V1/V2

C2 = 0.25 M ×  2.00 mL/50.00 mL

C2 = 0.01 M

The concentration of the final solution is 0.01 M.

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