Example 2

Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.

Given: identity of solute and volume and molarity of solution

Asked for: amount of solute in moles

Strategy:

Use either Equation 4.5 or Equation 4.6, depending on the units given in the problem.

Solution:

Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation 4.5 is more useful:

$$\text{moles NaOH }=V_{\text{L}}{\text{M}}_{\text{mol/L}}=\text{ (2}\text{.50 }\cancel{\text{L}}\text{)}\left(\frac{\text{0}\text{.100 mol}}{\cancel{\text{L}}}\right)=\text{ 0}\text{.250 mol NaOH}$$

Exercise

Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.

Answer: 41.6 mmol

Example 3

The solution in Figure 4.6 "Preparation of a Solution of Known Concentration Using a Solid Solute" 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,

$${\text{moles CoCl}}_{\text{2}}{\text{•2H}}_{\text{2}}\text{O }=\left(\frac{\text{10}\text{.0 }\cancel{\text{g}}}{\text{165}\text{.87 }\cancel{\text{g}}\text{/mol}}\right)=\text{ 0}\text{.0603 mol}$$

The volume of the solution in liters is

$$\text{volume }=\text{ 500 mL}\left(\frac{\text{1 L}}{\text{1000 }\cancel{\text{mL}}}\right)=\text{ 0}\text{.500 L}$$

Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is

$$\text{molarity }=\frac{\text{0}\text{.0603 mol}}{\text{0}\text{.500 L}}=\text{ 0}\text{.121 M }={\text{CoCl}}_{\text{2}}{\text{•H}}_{\text{2}}\text{O}$$

Exercise

The solution shown in Figure 4.7 "Preparation of 250 mL of a Solution of (NH" 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

Example 4

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:

$$\begin{array}{c}{V}_{\text{L}}{\text{M}}_{\text{mol/L}}=\text{ moles}\\ \text{500 }\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}\text{.310 mol glucose}}{\text{1}\cancel{\text{L}}}\right)=\text{ 0}\text{.155 mol glucose}\end{array}$$

B We then convert the number of moles of glucose to the required mass of glucose:

$$\text{mass of glucose }=\text{ 0}\text{.155 }\cancel{\text{mol glucose}}\left(\frac{\text{180}\text{.16 g glucose}}{\text{1 }\cancel{\text{mol glucose}}}\right)=\text{ 27}\text{.9 g glucose}$$

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

Example 5

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 4.7 to calculate the number of moles of glucose contained in 2500 mL of the solution:

$$\text{moles glucose }=\text{ 2500}\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}\text{.310 mol glucose}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 0}\text{.775 mol glucose}$$

B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:

$$\text{volume of stock soln }=\text{ 0}\text{.775 }\cancel{\text{mol glucose}}\left(\frac{\text{1 L}}{\text{3}\text{.00 }\cancel{\text{mol glucose}}}\right)=\text{ 0}\text{.258 L or 258 mL}$$

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 4.7 for V_s and substituting the appropriate values:

$$V_{\text{s}}=\frac{\text{(}{V}_{\text{d}}{\text{)(M}}_{\text{d}}\text{)}}{{\text{M}}_{\text{s}}}=\frac{\text{(2}\text{.500 L)(0}\text{.310 }\cancel{\text{M}}\text{)}}{\text{3}\text{.00 }\cancel{\text{M}}}=\text{ 0}\text{.258 L}$$

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

Example 6

What are the concentrations of all species derived from the solutes in these aqueous solutions?

1. 0.21 M NaOH
2. 3.7 M (CH₃)CHOH
3. 0.032 M In(NO₃)₃

Given: molarity

Asked for: concentrations

Strategy:

A Classify each compound as either a strong electrolyte or a nonelectrolyte.

B If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution.

Solution:

1. Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution:

   $$\text{NaOH(s)}\stackrel{{\text{H}}_{\text{2}}\text{O(l)}}{\to }{\text{Na}}^{+}\text{(aq)}+{\text{OH}}^{–}\text{(aq)}$$

   B Because each formula unit of NaOH produces one Na⁺ ion and one OH⁻ ion, the concentration of each ion is the same as the concentration of NaOH: [Na⁺] = 0.21 M and [OH⁻] = 0.21 M.

2. A The formula (CH₃)₂CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from Section 4.1 "Aqueous Solutions" that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes.

   B The only solute species in solution is therefore (CH₃)₂CHOH molecules, so [(CH₃)₂CHOH] = 3.7 M.

3. A Indium nitrate is an ionic compound that contains In³⁺ ions and NO₃⁻ ions, so we expect it to behave like a strong electrolyte in aqueous solution:

   $${\text{In(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(s)}\stackrel{{\text{H}}_{\text{2}}\text{O(l)}}{\to }{\text{In}}^{\text{3+}}\text{(aq)}+{\text{3NO}}_{\text{3}}{}^{–}\text{(aq)}$$

   B One formula unit of In(NO₃)₃ produces one In³⁺ ion and three NO₃⁻ ions, so a 0.032 M In(NO₃)₃ solution contains 0.032 M In³⁺ and 3 × 0.032 M = 0.096 M NO₃^–—that is, [In³⁺] = 0.032 M and [NO₃⁻] = 0.096 M.

Exercise

What are the concentrations of all species derived from the solutes in these aqueous solutions?

1. 0.0012 M Ba(OH)₂
2. 0.17 M Na₂SO₄
3. 0.50 M (CH₃)₂CO, commonly known as acetone

Answer:

1. [Ba²⁺] = 0.0012 M; [OH⁻] = 0.0024 M
2. [Na⁺] = 0.34 M; [SO₄²⁻] = 0.17 M
3. [(CH₃)₂CO] = 0.50 M