59.0 mL of a 1.60 M solution is diluted to a total volume of 268 mL. A 134-mL portion of that solution is diluted by adding 173 mL of water. What is the final concentration? Assume the volumes are additive?
59.0 mL of a 1.60 M solution is diluted to a total volume of 268 mL
The volume is increased by a factor of 268/59. So, the concentration is reduced by that factor.
59/268 * 1.60 = 0.3522M
Now the volume is increased by a factor of 307/134, so the concentration is reduced by that factor:
134/307 * .3522 = 0.154M
To find the final concentration of the solution, we need to use the formula for dilution:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution
Let's calculate the initial volume of the solution using the initial concentration and the final concentration.
C1V1 = C2V2
(1.60 M)(59.0 mL) = C2(268 mL)
To find V1, divide both sides of the equation by C1:
V1 = (C2V2) / C1
V1 = (1.60 M)(268 mL) / (59.0 mL)
V1 = 7.28 M
Now, we need to find the final volume of the diluted solution. We can add the volume of the initial solution (134 mL) with the volume of water added (173 mL):
Final Volume = 134 mL + 173 mL
Final Volume = 307 mL
Finally, we can use the formula to calculate the final concentration:
C1V1 = C2V2
(1.60 M)(7.28 mL) = C2(307 mL)
To find C2, divide both sides of the equation by V2:
C2 = (C1V1) / V2
C2 = (1.60 M)(7.28 mL) / (307 mL)
C2 ≈ 0.0378 M
Therefore, the final concentration of the solution after dilution is approximately 0.0378 M.