Diluting a Stock Solution for a Lab Experiment
A student needs to prepare 250.0 mL of a 0.150 M copper(II) sulfate solution for an experiment. The lab has a stock bottle of 1.20 M CuSO4 solution and distilled water. The student measures 31.3 mL of the stock solution using a graduated cylinder and adds it to a 250.0 mL volumetric flask, then fills to the mark with distilled water. After mixing, the student measures the absorbance of the solution in a spectrometer and finds it to be 0.450. A previously prepared calibration curve shows that absorbance = 3.00 × [CuSO4 concentration in M]. The student wonders if the dilution was accurate and if the solution is suitable for the experiment, which requires a concentration of 0.150 M ± 0.005 M.
1. Calculate the expected concentration of the diluted solution based on the volumes used. Is the student's dilution accurate?
Read the worked answer
Use the dilution equation: M1V1 = M2V2. M1 = 1.20 M, V1 = 31.3 mL, V2 = 250.0 mL. M2 = (1.20 M × 31.3 mL) / 250.0 mL = 0.15024 M ≈ 0.150 M. The expected concentration is 0.150 M, which is within the required range of 0.150 ± 0.005 M. So, based on volumes, the dilution is accurate.
2. Use the absorbance measurement to determine the actual concentration. Does it agree with the expected concentration? What might explain any discrepancy?
Read the worked answer
From the calibration curve: absorbance = 3.00 × concentration. Concentration = absorbance / 3.00 = 0.450 / 3.00 = 0.150 M. The measured concentration is 0.150 M, which matches the expected concentration. There is no significant discrepancy. Possible sources of small error could include reading the meniscus, temperature effects, or spectrometer calibration, but the result is within the acceptable range.
3. If the student had instead used 30.0 mL of stock solution, what would be the expected concentration? Would it still meet the requirement?
Read the worked answer
M2 = (1.20 M × 30.0 mL) / 250.0 mL = 0.144 M. The required range is 0.150 ± 0.005 M, i.e., 0.145 M to 0.155 M. 0.144 M is slightly below 0.145 M, so it would not meet the requirement. The student would need to adjust the volume or concentration.