Example 3.21: Volume of a Concentrated Solution Needed for Dilution
What volume of
1.59 M
\(\ce{KOH}\) is required to prepare
5.00 L of
0.100 M
\(\ce{KOH}\)?
Solution
We are given the concentration of a stock solution,
\(c_{\mathrm{1}}\), and the volume and concentration of the resultant diluted solution,
\(\ce{V2}\) and
\(c_{\mathrm{2}}\). We need to find the volume of the stock solution,
\(\ce{V1}\).
\(c_{\mathrm{1}}\) \(= 1.59\ \mathrm{M}\)
\(c_{\mathrm{2}}\) \(= 0.100\ \mathrm{M}\)
\(V_{\mathrm{2}}\) \(= 5.00\ \mathrm{L}\)
\(V_{\mathrm{1}}\) = ?
Since the concentration of the diluted solution
0.100 M is roughly one-sixteenth that of the stock solution (
1.59 M), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around
0.3 liters.
To obtain the accurate volume, we rearrange the dilution equation in order to isolate V1:
\(c_{\mathrm{1}} \cdot V_{\mathrm{1}} = c_{\mathrm{2}} \cdot V_{\mathrm{2}}\)
\(\ \ \ =5.00\ \mathrm{L} \cdot \dfrac{0.100\ \mathrm{M}}{1.59\ \mathrm{M}}\)
\(\ \ \ =5.00\ \mathrm{L} \cdot 0.06289\)
\(\ \ \ =0.314\ \mathrm{L}\)
Thus, we would need
0.314 L of the
1.59 M solution to prepare the desired solution. This result is consistent with our rough estimate.