What NaCl concentration results when 279 mL of a 0.640 M NaCl solution is mixed with 427 mL of a 0.490 M NaCl solution?
(.640)(.279) + (.490)(.427) = (x)(.279+.427)
x = .549 M
To find the resulting NaCl concentration when two solutions are mixed, you can use the formula:
C1V1 + C2V2 = C3V3
where C1 and C2 are the initial concentrations of the solutions, V1 and V2 are the volumes of the solutions, and C3 is the resulting concentration and V3 is the total volume after mixing.
Let's plug in the values given in the question:
C1 = 0.640 M
V1 = 279 mL
C2 = 0.490 M
V2 = 427 mL
V3 is the total volume after mixing, so V3 = V1 + V2 = 279 mL + 427 mL = 706 mL
Now, let's solve for C3:
C1V1 + C2V2 = C3V3
(0.640 M)(279 mL) + (0.490 M)(427 mL) = C3(706 mL)
177.76 + 209.23 = C3(706)
386.99 = 706C3
Divide both sides by 706 to solve for C3:
C3 = 386.99 / 706
C3 ≈ 0.548 M
Therefore, the resulting NaCl concentration when 279 mL of a 0.640 M NaCl solution is mixed with 427 mL of a 0.490 M NaCl solution is approximately 0.548 M.