Question 15.15: Calculating the Volume of Solution Needed to Supply a Given ......
Calculating the Volume of Solution Needed to Supply a Given Amount of an Ion
What volume, in milliliters, of a 0.300 M \textrm{Al} ( \textrm{NO}_3)_3 solution is needed to supply 0.782 mole of NO{_3} ^- ion?
The equation that governs the ions present in an aqueous \textrm{Al} ( \textrm{NO}_3)_3 solution is
\textrm{Al} ( \textrm{NO}_3)_3 (aq) \longrightarrow Al ^{3+} (aq) + 3 NO{_3} ^- (aq)
The extent of this dissociation is 100% as \textrm{Al} ( \textrm{NO}_3)_3 is a strong electrolyte (Sec. 15.2).
Step 1 The given quantity is 0.782 mole of NO{_3} ^- ion and the desired quantity is volume of \textrm{Al} ( \textrm{NO}_3)_3 solution.
0.782 mole NO{_3} ^- ion = ? mL \textrm{Al} ( \textrm{NO}_3)_3 solution
Step 2 This is moles-of-A to volume-of-solution B problem. The pathway, in terms of Figure 15.3, is
Step 3 The dimensional analysis setup for the calculation is
0.782 \cancel{{\textrm{NO}_3}^- \textrm{ ion}} \times \frac{1 \cancel{\textrm{mole Al} ( \textrm{NO}_3)_3}}{3 \cancel{{\textrm{moles NO}_3}^- \textrm{ ion}}} \times \frac{1000 \textrm{ mL Al}(\textrm{NO}_3)_3}{0.300 \cancel{\textrm{mole Al} ( \textrm{NO}_3)_3}}\quad \quad moles A → moles B → solution volume B
The numbers in the first conversion factor come from the dissociation equation and the numbers in the second conversion factor come from the given molarity of the solution.
Step 4 The result, obtained by combining the numerical factors, is
\frac{0.782 \times 1 \times 1000}{3 \times 0.300} \textrm{ mL Al} ( \textrm{NO}_3)_3 = 868.8889 \textrm{ mL Al} ( \textrm{NO}_3)_3 (calculator answer)
= 869 \textbf{ mL Al} ( \textbf{NO}_3)_3 (correct answer)
Answer Double Check: Is the magnitude of the answer reasonable? Yes. The 0.300 M \textrm{Al} ( \textrm{NO}_3)_3 is 0.900 molar in NO{_3} ^- ion. Thus, 1 liter (1000 mL) of solution would supply 0.900 M of NO{_3} ^- ion. A volume less than 1000 mL would be needed to supply the needed 0.782 mole of NO{_3} ^- ion, which is the case here.
