If you need to prepare 100 mL of a 1.0 x 10 4 M solution of potassium hydrogen p

Question

If you need to prepare 100 mL of a 1.0 x 104 M solution of potassium hydrogen phthalate (KHP) in two dilutions when given a 0.1000 M initial stock solution, which of the following procedures are correct?

(correct / incorrect) Pipet 10.0 mL of 0.1000 M stock solution into a 100-mL volumetric flask and dilute to the mark with deionized (DI) water, then pipet 5.0 mL of this diluted solution into another 100-mL volumetric flask and dilute to the mark with DI water.

(correct / incorrect)  Pipet 10.0 mL of 0.1000 M stock solution into a 50-mL volumetric flask and dilute to the mark with deionized (DI) water, then pipet 10.0 mL of this diluted solution into another 100-mL volumetric flask and dilute to the mark with DI water.

(correct / incorrect)  Pipet 1.0 mL of 0.1000 M stock solution into a 50-mL volumetric flask and dilute to the mark with deionized (DI) water, then pipet 10.0 mL of this diluted solution into another 100-mL volumetric flask and dilute to the mark with DI water.

(correct / incorrect) Pipet 1.0 mL of 0.1000 M stock solution into a 100-mL volumetric flask and dilute to the mark with deionized (DI) water, then pipet 10.0 mL of this diluted solution into another 100-mL volumetric flask and dilute to the mark with DI water.

Explanation / Answer

Fourth option is the correct choice.

We know the molarity after dilution can be calculated by using VM Before dilutioon = VM after dilution , as the moles of solute remain same.

option 1

Molarity after first dilution = 10x 0.1 /100 = 1.0x10-2 M

molarity after second dilution =1.0x10-2 x5/100 = 5x10-4 M

Thus incorrect

option 2

Molarity after first dilution = 10x 0.1 /50 = 2.0x10-2 M

molarity after second dilution =2.0x10-2 x10/100 = 2x10-3 M

Thus incorrect

option 3

Molarity after first dilution = 1x 0.1 /50 = 2.0x10-3 M

molarity after second dilution =2.0x10-3 x10/100 = 2x10-4 M

Thus incorrect

option 4

Molarity after first dilution = 1.0x 0.1 /100 = 1.0x10-3M

molarity after second dilution =1.0x10-3 x10 /100 = 1.0x10-4 M

Thus correct

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