Example 3.19: Determining the Concentration of a Diluted Solution

If 0.850 L of a 5.00-M solution of copper nitrate, \(\ce{Cu(NO3)2}\), is diluted to a volume of 1.80 L by the addition of water, what is the concentration of the diluted solution?

Solution

We are given the volume and concentration of a stock solution, \(\ce{V1}\) and \(c_{\mathrm{1}}\), and the volume of the resultant diluted solution, \(\ce{V2}\). We need to find the concentration of the diluted solution, \(c_{\mathrm{2}}\).

\(V_{\mathrm{1}}\) \(= 0.850\ \mathrm{L}\)


\(c_{\mathrm{1}}\) \(= 5.00\ \mathrm{M}\)


\(V_{\mathrm{2}}\) \(= 1.80\ \mathrm{L}\)


\(c_{\mathrm{2}}\) = ?


We thus rearrange the dilution equation in order to isolate c2:

\(c_{\mathrm{1}} \cdot V_{\mathrm{1}} = c_{\mathrm{2}} \cdot V_{\mathrm{2}}\)     

\(c_{\mathrm{2}} = \dfrac{c_{\mathrm{1}} \cdot V_{\mathrm{1}}}{V_{\mathrm{2}}}\)     

Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution’s concentration to be less than one-half 5 M. We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:

\(c_{\mathrm{2}}\) \(= c_{\mathrm{1}} \cdot \dfrac{V_{\mathrm{1}}}{V_{\mathrm{2}}}\)

\(\ \ \ =5.00\ \mathrm{M} \cdot \dfrac{0.850\ \mathrm{L}}{1.80\ \mathrm{L}}\)

\(\ \ \ =5.00\ \mathrm{M} \cdot 0.4722\)

\(\ \ \ =2.36\ \mathrm{M}\)


This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M).