Preparing a Copper(II) Sulfate Solution
At a regional science magnet school, a chemistry team must prepare 250.0 mL of 0.150 M copper(II) sulfate for a calorimetry investigation. The stock bottle is labeled 0.800 M CuSO4, and its certificate of analysis gives an uncertainty of ±0.005 M. Available glassware includes a 50.0 mL graduated cylinder with 1 mL scale divisions, a 25.0 mL volumetric pipet with ±0.03 mL tolerance, a 250.0 mL volumetric flask with ±0.12 mL tolerance, and a digital balance that reads to 0.01 g. One student proposes measuring the stock in the graduated cylinder because it is faster and “close enough.” The lab manual requires the final concentration to be within 2% of 0.150 M. The team must also report the final concentration with an appropriate number of significant figures and explain how each measurement choice affects uncertainty. The team has one attempt and must decide before pouring.
1. Calculate the volume of 0.800 M stock needed. Using propagation reasoning, decide whether the 50.0 mL graduated cylinder can meet the 2% tolerance, stating any assumption about its reading uncertainty.
Read the worked answer
Dilution requires $M_1V_1=M_2V_2$, so
$$V_1=\frac{(0.150\ \mathrm{M})(250.0\ \mathrm{mL})}{0.800\ \mathrm{M}}=46.875\ \mathrm{mL}\approx 46.9\ \mathrm{mL}.$$
The stock concentration has relative uncertainty $0.005/0.800=0.625%$. The volume uncertainty depends on what “1 mL scale divisions” means. If the student can read to ±0.5 mL, the relative volume uncertainty is $0.5/46.9=1.07%$, and the combined relative uncertainty is
$$\sqrt{0.625^2+1.07^2}\approx 1.24%,$$
which is within 2%. If the cylinder’s manufacturer tolerance is ±1 mL, the volume term becomes $1/46.9=2.13%$, and the combined uncertainty is about 2.22%, which fails. The judgment call is therefore conditional: the graduated cylinder is acceptable only if its actual tolerance is ±0.5 mL; otherwise the team should use a buret or weigh the stock by mass.
2. Report the final concentration with uncertainty and the correct number of significant figures. Explain why more digits should not be reported.
Read the worked answer
Using $V_1=46.9\ \mathrm{mL}$, the delivered concentration is
$$M_2=\frac{(0.800)(46.9)}{250.0}=0.15008\ \mathrm{M}.$$
The dominant relative uncertainty is about 1.1–1.3%, or roughly 0.002 M absolute. Report
$$0.150\ \mathrm{M}\pm 0.002\ \mathrm{M}.$$
The value 0.150 M has three significant figures, matching the 0.800 M stock and the 250.0 mL flask. Reporting 0.15008 M would falsely imply that the fourth and fifth digits are meaningful; the stock uncertainty alone makes the ten-thousandths place unreliable. The ±0.002 M interval communicates the real precision better than extra digits.
3. The team instead considers weighing solid CuSO4·5H2O directly. Calculate the mass needed for 250.0 mL of 0.150 M solution and judge whether the 0.01 g balance is precise enough.
Read the worked answer
Moles needed:
$$n=(0.150\ \mathrm{mol/L})(0.2500\ \mathrm{L})=0.03750\ \mathrm{mol}.$$
Using $M(\mathrm{CuSO_4\cdot 5H_2O})\approx 249.68\ \mathrm{g/mol}$,
$$m=(0.03750)(249.68)=9.363\ \mathrm{g}\approx 9.36\ \mathrm{g}.$$
The balance uncertainty is ±0.01 g, a relative uncertainty of $0.01/9.36=0.11%$, which is much smaller than the 2% tolerance. Numerically the balance is precise enough. The judgment call is that precision is not the only issue: the hydrate can be hygroscopic, so a nominally 9.36 g sample may contain extra water, and the solid may not be 100% pure. The team should dry or store the solid properly and, if high accuracy is required, standardize the solution rather than trusting the label alone.