Calculate the concentration (M)of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of sodium sulfide to a total volume of 250.0 mL.
A)0.175
B)4.37
C)0.525
D)0.350
E)0.874

Question

Quizplus · Accepted Answer

The concentration of sodium ions in the diluted solution can be calculated using the dilution formula $C_1V_1 = C_2V_2$, where $C_1$ and $V_1$ are the concentration and volume of the initial solution, and $C_2$ and $V_2$ are the concentration and volume of the diluted solution, respectively. Since sodium sulfide ($Na_2S$) dissociates into 2 sodium ions ($Na^+$) for every formula unit, the concentration of sodium ions in the original solution is $0.874 M \times 2 = 1.748 M$. Applying the dilution formula for the sodium ions: $1.748 M \times 50.0 mL = C_2 \times 250.0 mL$. Solving for $C_2$ gives $C_2 = 0.350 M$, which is the concentration of sodium ions in the diluted solution.

Calculate the Concentration (M)of Sodium Ions in a Solution Made