(Please Help) Solutions Topic - Chemistry

Question

1. A) Describe how you would make a stock solution of 1.25 mol/L potassium hydroxide, KOH

B) Describe how to dilute that solution to one with a concentration of 0.25 mol/L.

Describe Steps Carefully

Your have access to: solid potassium hydroxide, distilled water, Erlenmeyer flasks, volumetric

flasks, beakers, graduated cylinders, balances, scoopula, weighing boats, pipette, pipette pumps,

hot plates, glass stir rods, etc.

(your answer does not need to use all the materials )

Answer 1

1. Mol wt of Potassium Hydroxide is 56.11
2. 1.25 moles /L is 1.25 M
3. 56.11 grams of KOH in 1000 mL is 1M

5.611 grams of KOH in 100 mL is 1M

7.014 grams of KOH in 100 mL is 1.25 M [5.611 x 1.25 = 7014 grams]

Weigh 7.014 grams of KOH using weigh boat after tare weighing the weigh boat in the balance. Transfer carefully to a beaker. Use 25 ml pipette and transfer 25 mL distilled water to the beaker of KOH. Stir it and carefully transfer he solution to a 100 mL standard flask. Make it to 100 mL volume using distilled water and label it as 1.25M KOH. Stopper it and mix it.

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1. To prepare 0.25 M KOH from 1.25 M KOH

Use the formula:

V1 x M1 = V2 x M2. 0.25 M x 100 mL = V2 X 1.25 M

0.25 x 100/1.25 = 20.

Pipette 20 mL of 1.25 M of KOH solution to a fresh 100 mL standard flask and make it to volume with distilled water. Stopper it and mix it and label as 0.25 M KOH.