A stock solution is prepared by weighing out an appropriate portion of a pure solid or by measuring out an appropriate volume of a pure liquid, placing it in a suitable flask, and diluting to a known volume. Exactly how one measure’s the reagent depends on the desired concentration unit. For example, to prepare a solution with a known molarity you weigh out an appropriate mass of the reagent, dissolve it in a portion of solvent, and bring it to the desired volume. To prepare a solution where the solute’s concentration is a volume percent, you measure out an appropriate volume of solute and add sufficient solvent to obtain the desired total volume.
Describe how to prepare the following three solutions: (a) 500 mL of approximately 0.20 M NaOH using solid NaOH; (b) 1 L of 150.0 ppm Cu2+ using Cu metal; and (c) 2 L of 4% v/v acetic acid using concentrated glacial acetic acid (99.8% w/w acetic acid).
Solution
(a) Because the desired concentration is known to two significant figures, we do not need to measure precisely the mass of NaOH or the volume of solution. The desired mass of NaOH is
\[\frac {0.20 \text{ mol NaOH}} {\text{L}} \times \frac {40.0 \text{ g NaOH}} {\text{mol NaOH}} \times 0.50 \text{ L} = 4.0 \text{ g NaOH} \nonumber\]
To prepare the solution, place 4.0 grams of NaOH, weighed to the nearest tenth of a gram, in a bottle or beaker and add approximately 500 mL of water.
(b) Since the desired concentration of Cu2+ is given to four significant figures, we must measure precisely the mass of Cu metal and the final solution volume. The desired mass of Cu metal is
\[\frac {150.0 \text{ mg Cu}} {\text{L}} \times 1.000 \text{ M } \times \frac {1 \text{ g}} {1000 \text{ mg}} = 0.1500 \text{ g Cu} \nonumber\]
To prepare the solution, measure out exactly 0.1500 g of Cu into a small beaker and dissolve it using a small portion of concentrated HNO3. To ensure a complete transfer of Cu2+ from the beaker to the volumetric flask—what we call a quantitative transfer—rinse the beaker several times with small portions of water, adding each rinse to the volumetric flask. Finally, add additional water to the volumetric flask’s calibration mark.
(c) The concentration of this solution is only approximate so it is not necessary to measure exactly the volumes, nor is it necessary to account for the fact that glacial acetic acid is slightly less than 100% w/w acetic acid (it is approximately 99.8% w/w). The necessary volume of glacial acetic acid is
\[\frac {4 \text{ mL } \ce{CH3COOH}} {100 \text{ mL}} \times 2000 \text{ mL} = 80 \text{ mL } \ce{CH3COOH} \nonumber\]
To prepare the solution, use a graduated cylinder to transfer 80 mL of glacial acetic acid to a container that holds approximately 2 L and add sufficient water to bring the solution to the desired volume.
To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution.
\( Molarity of solution = dfrac{moles\: of\: solute}{Volume of solution} \tag{12.1.1}\)
Figure 12.1.1 illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the solvent is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always less than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in Figure 12.1.2, for some substances this effect can be significant, especially for concentrated solutions.
Figure 12.1.1 Preparation of a Solution of Known Concentration Using a Solid Solute
Figure 12.1.2 Preparation of 250 mL of a Solution of (NH4)2Cr2O7 in Water
The solute occupies space in the solution, so less than 250 mL of water are needed to make 250 mL of solution.
The solution in Figure 12.1.1 contains 10.0 g of cobalt(II) chloride dihydrate, CoCl2·2H2O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl2·2H2O?
Given: mass of solute and volume of solution
Asked for: concentration (M)
Strategy:
To find the number of moles of CoCl2·2H2O, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters.
Solution:
The molar mass of CoCl2·2H2O is 165.87 g/mol. Therefore,
\( moles\: CoCl_2 \cdot 2H_2O = \left( \dfrac{10.0 \: \cancel{g}} {165 .87\: \cancel{g} /mol} \right) = 0 .0603\: mol \)
The volume of the solution in liters is
\( volume = 500\: \cancel{mL} \left( \dfrac{1\: L} {1000\: \cancel{mL}} \right) = 0 .500\: L \)
Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is
\( molarity = \dfrac{0.0603\: mol} {0.500\: L} = 0.121\: M = CoCl_2 \cdot H_2O \)
Exercise
The solution shown in Figure 12.1.2 contains 90.0 g of (NH4)2Cr2O7 in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate?
Answer: (NH4)2Cr2O7 = 1.43 M
To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in Equation 12.1.1. We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example 12.1.2.
The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol.
Given: molarity, volume, and molar mass of solute
Asked for: mass of solute
Strategy:
A Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity.
B Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass.
Solution:
A We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution:
\( V_L M_{mol/L} = moles \)
\( 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .155\: mol\: glucose \)
B We then convert the number of moles of glucose to the required mass of glucose:
\( mass \: of \: glucose = 0.155 \: \cancel{mol\: glucose} \left( \dfrac{180.16 \: g\: glucose} {1\: \cancel{mol\: glucose}} \right) = 27.9 \: g \: glucose \)
Exercise
Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution.
Answer: 2.3 g NaCl
A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solution is a commercially prepared solution of known concentration and is often used for this purpose. Diluting a stock solution is preferred because the alternative method, weighing out tiny amounts of solute, is difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids.
The procedure for preparing a solution of known concentration from a stock solution is shown in Figure 12.1.3. It requires calculating the number of moles of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is therefore
\((V_s)(M_s) = moles\: of\: solute = (V_d)(M_d)\tag{12.1.2}\)
where the subscripts s and d indicate the stock and dilute solutions, respectively. Example 5 demonstrates the calculations involved in diluting a concentrated stock solution.
Figure 12.1.3 Preparation of a Solution of Known Concentration by Diluting a Stock Solution (a) A volume (Vs) containing the desired moles of solute (Ms) is measured from a stock solution of known concentration. (b) The measured volume of stock solution is transferred to a second volumetric flask. (c) The measured volume in the second flask is then diluted with solvent up to the volumetric mark [(Vs)(Ms) = (Vd)(Md)].
What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example 4?
Given: volume and molarity of dilute solution
Asked for: volume of stock solution
Strategy:
A Calculate the number of moles of glucose contained in the indicated volume of dilute solution by multiplying the volume of the solution by its molarity.
B To determine the volume of stock solution needed, divide the number of moles of glucose by the molarity of the stock solution.
Solution:
A The D5W solution in Example 4 was 0.310 M glucose. We begin by using Equation 12.1.2 to calculate the number of moles of glucose contained in 2500 mL of the solution:
\( moles\: glucose = 2500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .775\: mol\: glucose \)
B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:
\( volume\: of\: stock\: soln = 0 .775\: \cancel{mol\: glucose} \left( \dfrac{1\: L} {3 .00\: \cancel{mol\: glucose}} \right) = 0 .258\: L\: or\: 258\: mL \)
In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; only the amount of solvent has changed. The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M).
We could also have solved this problem in a single step by solving Equation 12.1.2 for Vs and substituting the appropriate values:
\( V_s = \dfrac{( V_d )(M_d )}{M_s} = \dfrac{(2 .500\: L)(0 .310\: \cancel{M} )} {3 .00\: \cancel{M}} = 0 .258\: L \)
As we have noted, there is often more than one correct way to solve a problem.
Exercise
What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)?
Answer: 16 mL