Multiple Choice
A 100.0 mL sample is prepared from a 250.0 mL stock solution by pipetting 5.00 mL of the stock solution and diluting it until it had a molarity of 2.5 * 10^-4 M. What is the molarity of the original stock solution?
6. Chemical Quantities & Aqueous Reactions
Dilutions
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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.
A
0.340 M
B
0.255 M
C
0.085 M
D
0.170 M
Verified step by step guidance1
Start by identifying the initial concentration and volume of the sodium sulfide solution. The initial concentration is 0.765 M and the initial volume is 25.0 mL.
Calculate the number of moles of sodium sulfide (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 sodium sulfide (Na2S) dissociates into two sodium ions (Na\(^+\)) and one sulfide ion (S\(^{2-}\)) in solution. Therefore, the concentration of sodium ions will be twice the concentration of sodium sulfide.
Determine the concentration of sodium ions in the initial solution by multiplying the concentration of sodium sulfide by 2.
Use the dilution formula \( C_1V_1 = C_2V_2 \) to find the final concentration of sodium ions after dilution. Here, \( C_1 \) is the initial concentration of sodium ions, \( V_1 \) is the initial volume, \( C_2 \) is the final concentration, and \( V_2 \) is the final volume (225 mL converted to L). Solve for \( C_2 \).
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