Question 4.9: Preparing a Solution by Dilution. A particular analytical ch......

Analyze

First, we calculate the number of moles K_{2}CrO_{4} that must be present in the final solution. Then, we calculate the volume of 0.250 M K_{2}CrO_{4} that contains this amount of K_{2}CrO_{4}.

Solve

First, calculate the amount of solute that must be present in the final solution.

? mol K_{2}CrO_{4} = 0.2500  L  soln × \frac{0.0100  mol   K_{2}CrO_{4}}{1  L  soln} = 0.00250  mol  K_{2}CrO_{4}

Second, calculate the volume of 0.250 M K_{2}CrO_{4} that contains 0.00250 mol K_{2}CrO_{4}.

? L soln = 0.00250 mol K_{2}CrO_{4} × \frac{1  L  soln}{0.250   mol  K_{2}CrO_{4}} = 0.0100 L soln

An alternative approach is to use equation (4.5). The known factors include the volume of solution to be prepared, (V_{f} = 250.0 mL) and the concentrations of the final (0.0100 M) and initial (0.250 M) solutions. We must solve for the initial volume, V_{i}. Note that although in deriving equation (4.5) volumes were expressed in liters, in applying the equation any volume unit can be used as long as we use the same unit for both V_{i}  and   V_{f} (milliliters in the present case). The term needed to convert volumes to liters would appear on both sides of the equation and cancel out.

c_{i}V_{i} = c_{f}V_{f}          (4.5)

V_{i} = V_{f} × \frac{c_{f}}{c_{i}} = 250.0 mL × \frac{0.0100   M}{0.250  M} = 10.0 mL

Assess

Let’s work the problem in reverse. If we dilute 10.0 mL of 0.250 M K_{2}CrO_{4} to 0.250 L with water, then the concentration of the diluted solution is 0.010 L × 0.250 mol L^{−1} / 0.250 L = 0.010 mol/L. This is the desired molarity; therefore, the answer, 10.0 mL of 0.250 M K_{2}CrO_{4}, is correct.