Multiple Choice
Calculate the concentration (M) of sodium ions in a solution made by diluting 25.0 mL of a 0.765 M solution of sodium sulfide (Na2S) to a total volume of 225 mL.
6. Chemical Quantities & Aqueous Reactions
Dilutions
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Calculate the concentration (M) of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of Na₂S to a total volume of 250.0 mL.
A
0.874 M
B
0.700 M
C
0.350 M
D
0.175 M
Verified step by step guidance1
Start by identifying the initial concentration and volume of the Na₂S solution. The initial concentration is 0.874 M and the initial volume is 50.0 mL.
Calculate the number of moles of Na₂S in the initial solution using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). Convert the volume from mL to L by dividing by 1000.
Recognize that each formula unit of Na₂S dissociates into two sodium ions (Na⁺) in solution. Therefore, the concentration of sodium ions will be twice the concentration of Na₂S.
Determine the concentration of Na₂S in the diluted solution using the dilution formula: \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( V_1 \) are the initial concentration and volume, and \( C_2 \) and \( V_2 \) are the final concentration and volume.
Calculate the final concentration of sodium ions by multiplying the concentration of Na₂S in the diluted solution by 2, since each Na₂S provides two Na⁺ ions.
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