Example 3.20: Volume of a Diluted Solution
What volume of
0.12 M
\(\ce{HBr}\) can be prepared from
11 mL (
0.011 L) of
0.45 M
\(\ce{HBr}\)?
Solution
We are given the volume and concentration of a stock solution,
\(\ce{V1}\) and
\(c_{\mathrm{1}}\), and the concentration of the resultant diluted solution,
\(c_{\mathrm{2}}\). We need to find the volume of the diluted solution,
\(\ce{V2}\).
\(c_{\mathrm{1}}\) \(= 0.45\ \mathrm{M}\)
\(V_{\mathrm{1}}\) \(= 0.011\ \mathrm{L}\)
\(c_{\mathrm{2}}\) \(= 0.12\ \mathrm{M}\)
\(V_{\mathrm{2}}\) = ?
Since the diluted concentration (
0.12 M) is slightly more than one-fourth the original concentration (
0.45 M), we would expect the volume of the diluted solution to be roughly four times the original volume, or around
44 mL.
To calculate the actual volume, we rearrange the dilution equation in order to isolate
\(\ce{V2}\).
\(c_{\mathrm{1}} \cdot V_{\mathrm{1}} = c_{\mathrm{2}} \cdot V_{\mathrm{2}}\)
Solving for the unknown volume yields:
\(V_{\mathrm{2}}\) \(= V_{\mathrm{1}} \cdot \dfrac{c_{\mathrm{1}}}{c_{\mathrm{2}}}\)
\(\ \ \ =0.011\ \mathrm{L} \cdot \dfrac{0.45\ \mathrm{M}}{0.12\ \mathrm{M}}\)
\(\ \ \ =0.011\ \mathrm{L} \cdot 3.75\)
\(\ \ \ =0.041\ \mathrm{L}\)
The volume of the
0.12 M solution is
0.041 L (
41 mL). The result is reasonable and compares well with our rough estimate.