The following questions all refer to an aqueous 8.41 M potassium hydroxide stock
ID: 912344 • Letter: T
Question
The following questions all refer to an aqueous 8.41 M potassium hydroxide stock solution. If 29.6 mL of the stock solution are diluted to 477 mL, what is the molarity (in mol/L) of the new solution? If 10.1mL of the stock solution are diluted to obtain a 0.750 M potassium hydroxide solution, what is the volume (in mL) of the new solution? Determine the volume (in mL) o the original stock solution that must be used to prepare 277 mL of a 0.692 M potassium hydroxide solution. Determine the volume of water (in mL) that was added to 21.2 mL of the stock solution to prepare a 0.152 M potassium hydroxide solution.Explanation / Answer
(1) We know,
V1S1 = V2S2 Here, V1 = Volume of the stock solution = 29.6 mL
S1 = Strength of the stock solution = 8.41 M = 8.41 mol /L
V2 = Volume of the new diluted solution = 477 mL
S2 = Strength of the new diluted solution = ?
Therefore,
S2 = V1S1 / V2
S2 = ( 29.6 mL) (8.41 mol/ L) / (477mL)
S2 = 0.522 mol/L
Thus, the molarity of the new diluted solution is 0.522 mol/L
(2) Here, V1 = Volume of the stock solution = 10.1 mL
S1 = Strength of the stock solution = 8.41 M
V2 = Volume of the new diluted solution = ?
S2 = Strength of the new diluted solution = 0.750 M
Therefore,
V2 = V1S1 / S2
V2 = (10.1 mL) (8.41 M) / (0.750 M)
V2 = 113 mL
The volume of the new diluted solution is 113 mL
(3) Here,
V1 = Volume of the stock solution = ?
S1 = Strength of the stock solution = 8.41 M
V2 = Volume of the new diluted solution = 277 mL
S2 = Strength of the new diluted solution = 0.692 M
Therefore,
V1 = V2S2 / S1
V1 = (277 mL) (0.692 M) / (8.41 M)
V1 = 22.8 mL
The required volume of the stock solution is 22.8 mL
(4) Here,
V1 = Volume of the stock solution = 21.2 mL
S1 = Strength of the stock solution = 8.41 M
V2 = Volume of the new diluted solution = ?
S2 = Strength of the new diluted solution = 0.152 M
Therefore,
V2 = V1S1 / S2
V2 = (21.2 mL) (8.41 M) / (0.152 M)
V2 = 1172.9 mL
Volume of the stock solution = 21.2 mL
Volume of the diluted new solution = 1172.9 mL
Volume of water added = (1172.9 - 21.2 ) mL = 1151.7 mL = 1152 mL
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