Solution A, while in a 1.0 cm cell, has an absorbance of 0.390 and a [Cu2+] = 0.0283 M. Then, 6.57 mL of solution A is diluted with pure water to 100.0 mL. What is the absorbance of this diluted solution?
I tried but I still don't get it. Can anyone please help! :( Thanks!
Solution A initial:
Absorbance = kc
Substitute A and c, calculate k.
Diluted solution:
new concn = 0.0283 x (6.57/100)
Substitute k and new c, calculate Absorbance.
To find the absorbance of the diluted solution, we can use the Beer-Lambert Law, which relates the concentration of a solute to its absorbance. The formula is A = ε * l * c, where A is the absorbance, ε is the molar absorptivity (constant), l is the path length of the solution cell, and c is the concentration of the solute.
Given:
A1 = 0.390 (absorbance of solution A)
[Cu2+]1 = 0.0283 M (concentration of solution A)
V1 = 6.57 mL (volume of solution A)
V2 = 100.0 mL (final volume after dilution)
First, let's find the concentration of the diluted solution:
[Cu2+]2 = [Cu2+]1 * (V1 / V2)
[Cu2+]2 = 0.0283 M * (6.57 mL / 100.0 mL)
[Cu2+]2 = 0.00185591 M
Next, we need to find the new absorbance (A2) using the concentration of the diluted solution:
A2 = A1 * (c2 / c1)
A2 = 0.390 * (0.00185591 M / 0.0283 M)
A2 ≈ 0.02557
Therefore, the absorbance of the diluted solution is approximately 0.02557.