75.0 mL of a 1.20 M solution is diluted to a total volume of 248 mL. A 124-mL portion of that solution is diluted by adding 111 mL of water. What is the final concentration? Assume the volumes are additive.
1.20 x (75.0/248) x [124/(124+111)] = ?
To find the final concentration, we can use the formula:
\(C_1V_1 = C_2V_2\)
where:
\(C_1\) = initial concentration of the solution
\(V_1\) = initial volume of the solution
\(C_2\) = final concentration of the solution
\(V_2\) = final volume of the solution
Given:
\(C_1 = 1.20 \, \text{M}\) (initial concentration)
\(V_1 = 75.0 \, \text{mL}\) (initial volume)
\(V_2 = 248 \, \text{mL}\) (final volume)
First, we need to find the final concentration after diluting the 75.0 mL of 1.20 M solution to a total volume of 248 mL. We can rearrange the formula as:
\(C_2 = \frac{{C_1 \times V_1}}{{V_2}}\)
Substituting the given values:
\(C_2 = \frac{{1.20 \, \text{M} \times 75.0 \, \text{mL}}}{{248 \, \text{mL}}}\)
\(C_2 = 0.3636 \, \text{M}\)
Now that we have found the final concentration after diluting the 75.0 mL of solution to 248 mL, we need to dilute a 124 mL portion of that solution by adding 111 mL of water. Since volumes are additive, the final volume is:
\(V_2 = 124 \, \text{mL} + 111 \, \text{mL} = 235 \, \text{mL}\)
Using the formula \(C_1V_1 = C_2V_2\), we can find the final concentration (\(C_2\)) after diluting the 124 mL portion:
\(C_2 = \frac{{C_1 \times V_1}}{{V_2}}\)
Substituting the values:
\(C_2 = \frac{{0.3636 \, \text{M} \times 124 \, \text{mL}}}{{235 \, \text{mL}}}\)
\(C_2 = 0.1922 \, \text{M}\)
Therefore, the final concentration after diluting the 124 mL portion with 111 mL of water is 0.1922 M.