Question

You have a stock solution certified by a manufacturer to contain $\mathbf{150}\mathbf{.}\mathbf{0}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{3}{\mathbf{\mu gSO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{}\mathbf{/}\mathbf{mL}$. You would like to dilute it by a factor of 100 to obtain $\mathbf{1}\mathbf{.}\mathbf{500}\mathit{\mu }\mathit{g}\mathbf{/}\mathit{m}\mathit{L}$. Two possible methods of dilution are stated below. For each method, calculate the resulting uncertainty in concentration. Use manufacturer's tolerances in Tables 2-3 and 2-4 for uncertainties. Explain why one method is more precise than the other.

(a) Dilute 10.00mL up to 100mL with a transfer pipet and volumetric flask. Then take 10.00mL of the dilute solution and dilute it again to 100mL.

(b) Dilute 1.000mL up to 100mL with a transfer pipet and volumetric flask.

Short answer

Uncertainty of first process will be 0.084 mol/L and the uncertainty of second process will be 0.0012 mol/L.

Because of error propagation, the second process is more precise. The uncertainty of each step will add up, thus there will be more uncertainty as the number of steps increases in the first process.

Step-by-step solution

Given

(a)

Process 1

(A)10 mL.... (B)100 mL.... (C) 10 mL.... (D)100 mL

To find percentage uncertainty

$$\begin{gathered} \mathrm{A}. \\ \frac{0.02}{10}\times 100 \\ =0.2\% \\ \mathrm{B}. \\ \frac{0.08}{100}\times 100 \\ =0.08\% \\ \mathrm{C}. \\ \frac{0.02}{10}\times 100 \\ =0.2\% \\ \mathrm{D}. \\ \frac{0.08}{100}\times 100 \\ =0.08\% \end{gathered}$$

To find total percentage uncertainty

Total percentage uncertainty

=0.2 + 0.08 + 0.2 + 0.08

=0.56 %

To find absolute uncertainty in concentration

Absolute uncertainty in concentration

$$\begin{gathered} =\frac{0.56}{100}\times 1.500 \\ =0.084\mathrm{mol}/\mathrm{L} \end{gathered}$$

To find percentage uncertainty for process 2

Process 2

1mL....100mL

Percentage uncertainty

$$\begin{gathered} =\frac{0.08}{100}\times 100 \\ =0.08\% \end{gathered}$$

To find absolute uncertainty in concentration

Absolute uncertainty

$$\begin{gathered} =\frac{0.08}{100}\times 1.500 \\ =0.0012\mathrm{mol}/\mathrm{L} \end{gathered}$$

Final conclusion

Because of error propagation, the second process is more precise. The uncertainty of each step will add up, thus there will be more uncertainty as the number of steps increases in the first process.