9.2 Concentration

Learning Objectives

  1. Express the amount of solute in a solution in various concentration units.
  2. Use molarity to determine quantities in chemical reactions.
  3. Determine the resulting concentration of a diluted solution.

To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms whose meanings depend on various factors.

Solubility

There is usually a limit to how much solute will dissolve in a given amount of solvent. This limit is called the solubility of the solute. Some solutes have a very small solubility, while other solutes are soluble in all proportions. Table 9.2 “Solubilities of Various Solutes in Water at 25°C “ lists the solubilities of various solutes in water. Solubility varies with temperature, so it is important to include the temperature at which the solubility was determined.  Most solids have a higher solubility in water at higher temperatures. For example, while 91 g of glucose will dissolve in 100 mL of water at 25oC, 357 g will dissolve at 70oC.

Table 9.2 Solubilities of Various Solutes in Water at 25°C

Substance Solubility (g in 100 mL of H2O)
AgCl(s) 0.019
C6H6(ℓ) (benzene) 0.178
CH4(g) 0.0023
CO2(g) 0.150
CaCO3(s) 0.058
CaF2(s) 0.0016
Ca(NO3)2(s) 143.9
C6H12O6 (glucose) 90.9
KBr(s) 67.8
MgCO3(s) 2.20
NaCl(s) 36.0
NaHCO3(s) 8.41
C12H22O11 (sucrose) 211.4

If a solution contains so much solute that its solubility limit is reached, the solution is said to be saturated. If a solution contains less solute than the solubility limit, it is unsaturated. For some substances, a supersaturated solution can be prepared by dissolving as much solute as possible at a higher temperature, then lowering the temperature.  Less solute should dissolve at the lower temperature, but the excess solute might not precipitate right away.  Thus the solution is supersaturated, but not stable. Eventually the excess solute will crystallize with a release of energy as the mixture achieves a more stable state.  Honey is a supersaturated solution which bees form by using their wings to fan nectar, a dilute sugar solution, to drive off much of the water, forming supersaturated honey.  The excess sugar in honey sometimes recrystallizes, making the honey cloudy or grainy.  It can be restored to a clear state by heating it.

In contrast to increased solubility of most solids at higher temperatures, the solubility of gases in liquid water decreases as temperature increases.  The solubility of O2 is 8.2 mg O2/L of H2O at 25oC but only 6.0 mg O2/L of H2O at 45oC, both at a pressure of 1 atm.    Pressure must be specified because it also influences the solubility of gases; more gas dissolves when the pressure is higher.   At 0.5 atm, 4.1 mg of O2 dissolves per L of H2O, but at 2 atm, the solubility of O2 is 16.5 mg/L, both at a temperature of 25 oC.   When pressure is suddenly changed, such as when you open a can of soda, excess gas can bubble out of solution uncontrollably.

Note

Reusable hot packs contain a supersaturated solution of sodium acetate.  When recrystallization is initiated using a clicker, the excess sodium acetate (NaC2H3O2) comes out of solution and heat is emitted.  The hot pack can be made ready for reuse by placing it in a pan of boiling water until the sodium acetate dissolves.

Most solutions we encounter are unsaturated, so knowing the solubility of the solute does not accurately express the amount of solute in these solutions. There are several common ways of specifying the concentration of a solution. 

Percent Composition

There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100:

[latex]\text{%m/m}=\frac{\text{mass solute}}{\text{mass solution}}\times{100}[/latex]

Each mass must be expressed in the same units to determine the proper concentration.  The mass of the solution may be given, or it may be necessary to calculate it by adding the masses of the solute and solvent.

Example 2

A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution?

Solution

We can substitute the quantities given in the equation for mass/mass percent:

[latex]\text{%m/m}=\frac{\text{mass solute}}{\text{mass solution}}\times{100}=\frac{36.5\text{ g}}{355\text{ g}}\times{100}=10.3\text{%}[/latex]

Skill-Building Exercise

  1. A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution?

For liquids, volumes are relatively easy to measure, so the concentration of a liquid in solution can be expressed as a volume/volume percent (% v/v): the volume of a solute divided by the volume of a solution times 100:

[latex]\text{%v/v}=\frac{\text{volume solute}}{\text{volume solution}}\times{100}[/latex]

As with mass/mass %, the units of the solute and the solution must be the same.  Both the volume of the solute and the volume of the solution must bespecified because volumes are not additive.  For example, a mixture made from 50 mL of water and 50 mL of ethanol has a volume of about 95 ml because the molecules of the two substances are attracted to each other and enter the spaces between each other’s molecules.

A hybrid concentration unit, mass/volume percent (% m/v), is commonly used for intravenous (IV) fluids (Figure 9.2 “Mass/Volume Percent”). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100:

[latex]\text{%m/v}=\frac{\text{grams solute}}{\text{mL solution}}\times{100}[/latex]

Unlike %m/m and %m/v, the units do not cancel and m/v% is not a true percentage.  Never the less, m/v % is widely used in medical settings because it is easy to deliver a certain mass of a medication or other solute by delivering a certain volume of solution.

Each percent concentration can be used to produce a conversion factor between the amount of solute, the amount of solution, and the percent. Furthermore, given any two quantities in any percent composition, the third quantity can be calculated, as the following example illustrates.  While it is possible to substitute known quantities into the percentage equation and solve for the missing quantity, it is easier to interpret percentage and use it as a conversion factor.  x % implies the following pair of conversion factors:

[latex]\frac{\text{x g or mL of solute}}{100\text{ g or mL of solution}}[/latex]   AND   [latex]\frac{100\text{ g or mL of solution}}{\text{ x g or mL of solute}}[/latex]    As always, use the conversion factor that allows the units to cancel.  Make sure to include a descriptor with the units, such as mL solution or g solvent- cancellation must include the unit and the descriptor.  Note that the 100 is placed there by definition of percent, so it does not limit the sig figs of the answer.

Example 3

A sample of 45.0% v/v solution of ethanol (C2H5OH) in water has a volume of 115 mL. What volume of ethanol solute does the sample contain?

Solution

There are two pieces of information given, the volume and the percent.  When percent is given, it will be used as the conversion factor, so begin with the volume.

[latex]115\cancel{\text{ mL solution}}\times{\frac{45.0\text{ mL ethanol}}{100\cancel{\text{ mL solution}}}}=51.8\text{ mL ethanol}[/latex]

The answer makes sense because a bit more than 100 mL of solution should contain a bit more than 45 g of ethanol.

Note

The highest concentration of ethanol that can be obtained normally is 95% ethanol, which is actually 95% v/v.

Skill-Building Exercise

  1. What volume of a 12.75% m/v solution of glucose (C6H12O6) in water is needed to obtain 50.0 g of C6H12O6?

Skill-Building Exercise

  1. The chlorine bleach that you might find in your laundry room is typically composed of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mL of solution. What is the mass/volume percent of the bleach?

In addition to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm) and parts per billion (ppb). Both of these units can be based on either mass or volume and are defined as follows:

[latex]\text{ppm}=\frac{\text{g or mL solute}}{\text{g or mL solution}}\times1,000,000[/latex]    and     [latex]\text{ppb}=\frac{\text{g or mL solute}}{\text{g or mL solution}}\times1,000,000,000[/latex]

Note

Similar to parts per million and parts per billion, related units include parts per thousand (ppth) and parts per trillion (ppt).

Concentrations of trace elements in the body—elements that are present in extremely low concentrations but are nonetheless necessary for life—are commonly expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon’s Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm.

Note

In aqueous solutions, 1 ppm is essentially equal to 1 mg/L, and 1 ppb is equivalent to 1 µg/L.

Example 4

If the concentration of cobalt in a human body is 21 ppb, what mass in grams of Co is present in a body having a mass of 70.0 kg?

Solution

The starting point is the simpler value given, the body mass of 70.0 kg.

A concentration of 21 ppb means “21 g of solute per 1,000,000,000 g (or 109g)  of body mass.”   Written as a pair of conversion factors::

[latex]\frac{21\text{ g Co}}{10^9\text{ g body mass}}[/latex]   AND   [latex]\frac{19^9\text{ g body mass}}{21\text{ g Co}}[/latex]

But before applying one of these conversion factors, the body mass must be converted from kg to g:

[latex]70.0\cancel{\text{ kg}}\times{\frac{1000\text{ g}}{1\cancel{\text{kg}}}}=7.00\times{10^4}\text{ g}[/latex]

Now we determine the amount of Co:

[latex]7.00\times{10^4}\cancel{\text{ g body mass}}\times{\frac{21\text{ g Co}}{10^9\cancel{\text{ g body mass}}}}=0.0015\text{ g Co}[/latex]

This is only 1.5 mg.

Skill-Building Exercise

  1. An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in parts per million?

Molarity

Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions that take place in solution. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.

Molarity is defined as the number of moles of a solute dissolved per liter of solution:

[latex]\text{molarity}=\frac{\text{number of moles solute}}{\text{number of liters solution}}[/latex]

Notice that the units are complex, mol/L, often abbreviated to just M.  Also note that this is not a percentage so there is no multiplication by 100.

For example, for 1.5 mol of NaCl dissolved in 0.500 L of solution, calculate the molarity as follows:

[latex]\text{molarity}=\frac{\text{number of moles solute}}{\text{number of liters solution}}=\frac{1.5\text{ mol NaCl}}{0.500\text{ L solution}}=3.0\text{ mol/L}=3.0\text{ M NaCl}[/latex]

Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example.

Example 5

What is the molarity of an aqueous solution of 25.0 g of NaOH in 750. mL?

Solution

Before substituting these quantities into the definition of molarity, convert them to the proper units. The mass of NaOH must be converted to moles of NaOH using  molar mass of NaOH as a conversion factor.  The molar mass of NaOH is 40.00 g/mol.

[latex]25.0\cancel{\text{ g NaOH}}\times{\frac{1 \text{ mol NaOH}}{40.00\cancel{\text{g NaOH}}}}=0.625\text{ mol NaOH}[/latex]

Next, convert the volume units from milliliters to liters:

[latex]750.\cancel{\text{ mL}}\times{\frac{1\text{ L}}{1000\cancel{\text{ mL}}}}=0.750\text{ L}[/latex]

Now that the quantities are expressed in the proper units, substitute them into the definition of molarity:

[latex]\text{molarity}=\frac{\text{number of moles solute}}{\text{number of liters solution}}=\frac{0.625\text{ mol NaOH}}{0.750\text{ L solution}}=3.0\text{ mol/L}=0.833\text{ M NaOH}[/latex]

Skill-Building Exercise

  1. If a 350 mL cup of coffee contains 0.150 g of caffeine (C8H10N4O2), what is the molarity of this caffeine solution?

When molarity is given, it can be interpreted as a pair of conversion factors that can be used to convert between moles of solute and liters of solution.  A molariy of x :

[latex]\frac{\text{x moles solute}}{1\text{ L solution}}[/latex]    and     [latex]\frac{1\text{ L solution}}{\text{x moles solute}}[/latex]

Note that the 1 L is part of the definition of molarity, so it does not limit the number of sig figs in the answer.

Example 6

What volume of a 0.0753 M solution of dimethylamine [(CH3)2NH] is needed to obtain 0.450 mol of the compound?

Solution

Begin with the given that has simpler units, 0.450 mol.  Remember that M means mols/L, a complex unit, so it will be used to create a conversion factor.

[latex]0.450\text{ mol dimethylamine}\times{\frac{1\text{ L solution}}{0.0753\text{ moles dimethylamine}}}=5.98\text{ L solution}[/latex]

Example 7

Ethylene glycol (C2H6O2) is mixed with water to make auto engine coolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueous solution?  The molar mass of C2H6O2 is 62.08 g/mol.

Three bits of information are provided, molarity and molar mass which both have complex units and are likely to be used as conversion factors, and volume which has a simple unit and thus is the starting point.

[latex]5.00\cancel{\text{ L solution}}\times{\frac{6.00\cancel{\text{ mol ethylene glycol}}}{1\cancel{\text{ L solution}}}}\times{\frac{62.08\text{ g ethylene glycol}}{1\cancel{\text{ mol ethylene glycol}}}}=1860\text{ g ethylene glycol}[/latex]

Note

Dimethylamine has a “fishy” odor. In fact, organic compounds called amines cause the odor of decaying fish. (For more information about amines, see Chapter 15 “Organic Acids and Bases and Some of Their Derivatives”, Section 15.1 “Functional Groups of the Carboxylic Acids and Their Derivatives” and Section 15.11 “Amines: Structures and Names” through Section 15.13 “Amines as Bases”.)

Skill-Building Exercise

  1. What volume of a 0.0902 M solution of formic acid (HCOOH) is needed to obtain 0.888 mol of HCOOH?

  2. Acetic acid (HC2H3O2) is the acid in vinegar. How many grams of HC2H3O2 are in 0.565 L of a 0.955 M solution?

Using Molarity in Stoichiometry Problems

Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit.

The general steps for performing stoichiometry problems using molarities of solutions  are shown in Figure 9.3 “Diagram of Steps for Using Molarity in Stoichiometry Calculations”. You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure 9.3 “Diagram of Steps for Using Molarity in Stoichiometry Calculations” indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end.

image

Figure 9.3 Diagram of Steps for Using Molarity in Stoichiometry Calculations

In itself, each step is a straightforward conversion. It is the combination of the steps that is a powerful quantitative tool for problem solving.

Example 8

What volume of a 2.75 M HCl solution is needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows:

HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq)

Solution

This problem gives grams of NaOH, so converting to moles NaOH using NaOH’s molar mass is the first step.  Next is the central part of any stoichiometry problem, relating moles of one substance to moles of a different substance using the mole-to-mole ratio based on coefficients from the balanced equation. Finally, the volume of HCl solution is the unknown, so the molarity of the HCl solution becomes the conversion factor between g HCl and volume of HCl solution.

[latex]185\cancel{\text{ g NaOH}}\times{\frac{1\cancel{\text{ mol NaOH}}}{40.00\cancel{\text{ g NaOH}}}}\times{\frac{1\cancel{\text{ mol HCl}}}{1\cancel{\text{ mol NaOH}}}}\times{\frac{1\text{ L HCl solution}}{2.75\cancel{\text{mol HCl}}}}=1.68\text{ L HCl solution}[/latex]

The answer could also be expressed in mL by multiplying by 1000, so 1680 mL.

Skill-Building Exercise

  1. How many milliliters of a 1.04 M H2SO4 solution are needed to react with 98.5 g of Ca(OH)2? The balanced chemical equation for the reaction is as follows:

    H2SO4(aq) + Ca(OH)2(s) → 2H2O(ℓ) + CaSO4(aq)

Concentrations of Substances in Bodily Fluids

Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins. Table 9.3 “Approximate Concentrations of Various Solutes in Some Solutions in the Body*” lists the typical concentrations of some of these solutes.

Table 9.3 Approximate Concentrations of Various Solutes in Some Solutions in the Body*

Solution Solute Concentration (M)
blood plasma Na+ 0.138
K+ 0.005
Ca2+ 0.004
Mg2+ 0.003
Cl− 0.110
HCO3− 0.030
stomach acid HCl 0.10
urine NaCl 0.15
PO43− 0.05
NH2CONH2 (urea) 0.30
*Note: Concentrations are approximate and can vary widely.

Looking Closer: The Dose Makes the Poison

Why is it that we can drink 1 qt of water when we are thirsty and not be harmed, but if we ingest 0.5 g of arsenic, we might die? There is an old saying: the dose makes the poison. This means that what may be dangerous in some amounts may not be dangerous in other amounts.

Take arsenic, for example. Some studies show that arsenic deprivation limits the growth of animals such as chickens, goats, and pigs, suggesting that arsenic is actually an essential trace element in the diet. Humans are constantly exposed to tiny amounts of arsenic from the environment, so studies of completely arsenic-free humans are not available; if arsenic is an essential trace mineral in human diets, it is probably required on the order of 50 ppb or less. A toxic dose of arsenic corresponds to about 7,000 ppb and higher, which is over 140 times the trace amount that may be required by the body. Thus, arsenic is not poisonous in and of itself. Rather, it is the amount that is dangerous: the dose makes the poison.

Similarly, as much as water is needed to keep us alive, too much of it is also risky to our health. Drinking too much water too fast can lead to a condition called water intoxication, which may be fatal. The danger in water intoxication is not that water itself becomes toxic. It is that the ingestion of too much water too fast dilutes sodium ions, potassium ions, and other salts in the bloodstream to concentrations that are not high enough to support brain, muscle, and heart functions. Military personnel, endurance athletes, and even desert hikers are susceptible to water intoxication if they drink water but do not replenish the salts lost in sweat. As this example shows, even the right substances in the wrong amounts can be dangerous!

Expressing Concentration of Ionic Substances in Equivalents

Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq). One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol/L of Na+(aq) is also 1 Eq/L because sodium has a 1+ charge. A 1 mol/L solution of Ca2+(aq) ions has a concentration of 2 Eq/L because calcium has a 2+ charge. Dilute solutions may be expressed in milliequivalents (mEq)—for example, human blood plasma has a total concentration of about 150 mEq/L. (For more information about the ions present in blood plasma, see Chapter 3 “Ionic Bonding and Simple Ionic Compounds”, Section 3.3 “Formulas for Ionic Compounds”.)

Dilution

When solvent is added to dilute a solution, the volume of the solution changes, but the amount of solute does not change. Before dilution, the amount of solute was equal to its original concentration (Ci) times its original volume (Vi).

amount of substance = Ci × Vi

After dilution, the same amount of solute is equal to the final concentration (Cf) times the final volume (Vf):

amount of substance = Cf × Vf

Since the amount of substance does not change during dilution:

Ci × Vi= Cf × Vf

Any units of concentration and volume can be used, as long as both concentrations and both volumes have the same unit.  Three of the values must be given in the problem so that the fourth value can be determined algebraically.

Example 9

A 125 mL sample of 0.900 M NaCl is diluted to 1,125 mL. What is the final concentration of the diluted solution?

Solution

The numbers given in the problem are Vi, Ci, and Vf, respectively.  Cf is the unknown.  To solve for Cf, divide both sides by Vf.

[latex]\frac{\text{C}_\text{i}\times{\text{V}_\text{i}}}{\text{V}_\text{f}}=\frac{\text{C}_\text{f}\times{\text{V}_\text{f}}}{\text{V}_\text{f}}[/latex]   so after cancelling   [latex]\text{C}_\text{f}=\frac{\text{C}_\text{i}\times{\text{V}_\text{i}}}{\text{V}_\text{f}}[/latex]

Filling in the actual values, [latex]\text{C}_\text{f}=\frac{0.900\text{ M}\times{125\text{ mL}}}{1125\text{M}}=0.100\text{ M}[/latex]

The answer is reasonable because dilution produces a larger volume with a smaller concentration.

Skill-Building Exercise

  1. A nurse uses a syringe to inject 5.00 mL of 0.550 M heparin solution (heparin is an anticoagulant drug) into a 250 mL IV bag, for a final volume of 255 mL. What is the concentration of the resulting heparin solution?

Concept Review Exercises

  1. What are some of the units used to express concentration?

  2. Distinguish between the terms solubility and concentration.

Answers

  1. % m/m, % m/v, ppm, ppb, molarity, and Eq/L (answers will vary)

  2. Solubility is typically a limit to how much solute can dissolve in a given amount of solvent. Concentration is the quantitative amount of solute dissolved at any concentration in a solvent.

Key Takeaways

  • Various concentration units are used to express the amounts of solute in a solution.
  • Concentration units can be used as conversion factors in stoichiometry problems.
  • New concentrations can be easily calculated if a solution is diluted.

Exercises

  1. Define solubility. Do all solutes have the same solubility?

  2. Explain why the terms dilute or concentrated are of limited usefulness in describing the concentration of solutions.

  3. Calculate the mass/mass percent of a saturated solution of NaCl. Use the data from Table 9.2 “Solubilities of Various Solutes in Water at 25°C “.   H2O has a density of 1.00 g/mL so the 100.0 mL of water is also 100.0 g of water.

  4. Calculate the mass/mass percent of a saturated solution of MgCO3 Use the data from Table 9.2 “Solubilities of Various Solutes in Water at 25°C “.   H2O has a density of 1.00 g/mL so the 100.0 mL of water is also 100.0 g of water,

  5. Only 0.203 mL of benzene, C6H6 will dissolve in 100.000 mL of H2O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of benzene in water.

  6. Only 35 mL of aniline (C6H5NH2) will dissolve in 1,000 mL of H2O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of aniline in water.

  7. A solution of ethyl alcohol (C2H5OH) in water has a concentration of 20.56% v/v. What volume of C2H5OH is present in 255 mL of solution?

  8. What mass of KCl is present in 475 mL of a 1.09% m/v aqueous solution?

  9. The average human body contains 5,830 g of blood. What mass of arsenic is present in the body if the amount in blood is 0.55 ppm?

  10. The Occupational Safety and Health Administration has set a limit of 200 ppm as the maximum safe exposure level for carbon monoxide (CO). If an average breath has a mass of 1.286 g, what is the maximum mass of CO that can be inhaled at that maximum safe exposure level?

  11. What is the molarity of 0.500 L of a potassium chromate solution containing 0.0650 mol of K2CrO4?

  12. What is the molarity of 4.50 L of a solution containing 0.206 mol of urea [(NH2)2CO]?

  13. What is the molarity of a 2.66 L aqueous solution containing 56.9 g of NaBr?

  14. If 3.08 g of Ca(OH)2 is dissolved in enough water to make 0.875 L of solution, what is the molarity of the Ca(OH)2?

  15. What mass of HCl is present in 825 mL of a 1.25 M solution?

  16. What mass of isopropyl alcohol (C3H8O) is dissolved in 2.050 L of a 4.45 M aqueous C3H8O solution?

  17. What volume of 0.345 M NaCl solution is needed to obtain 10.0 g of NaCl?

  18. How many milliliters of a 0.0015 M cocaine hydrochloride (C17H22ClNO4) solution is needed to obtain 0.010 g of the solute?

  19. Aqueous calcium chloride reacts with aqueous silver nitrate according to the following balanced chemical equation:

    CaCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Ca(NO3)2(aq)

    How many moles of AgCl(s) are made if 0.557 L of 0.235 M CaCl2 react with excess AgNO3? How many grams of AgCl are made?

  20. Sodium bicarbonate (NaHCO3) is used to react with acid spills. The reaction with sulfuric acid (H2SO4) is as follows:

    2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2H2O(ℓ) + 2CO2(g)

    If 27.6 mL of a 6.25 M H2SO4 solution were spilled, how many moles of NaHCO3 would be needed to react with the acid? How many grams of NaHCO3 is this?

  21. The fermentation of glucose to make ethanol and carbon dioxide has the following overall chemical equation:

    C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)

    If 1.00 L of a 0.567 M solution of C6H12O6 were completely fermented, what would be the resulting concentration of the C2H5OH solution? How many moles of CO2 would be formed? How many grams is this? If each mole of CO2 had a volume of 24.5 L, what volume of CO2 is produced?

  22. Aqueous sodium bisulfite gives off sulfur dioxide gas when heated:

    2NaHSO3(aq) → Na2SO3(aq) + H2O(ℓ) + SO2(g)

    If 567 mL of a 1.005 M NaHSO3 solution were heated until all the NaHSO3 had reacted, what would be the resulting concentration of the Na2SO3 solution? How many moles of SO2 would be formed? How many grams of SO2 would be formed? If each mole of SO2 had a volume of 25.78 L, what volume of SO2 would be produced?

  23. What is the concentration of a 1.0 M solution of K+(aq) ions in equivalents/liter?

  24. What is the concentration of a 1.0 M solution of SO42−(aq) ions in equivalents/liter?

  25. A solution having initial concentration of 0.445 M and initial volume of 45.0 mL is diluted to 100.0 mL. What is its final concentration?

  26. A 50.0 mL sample of saltwater that is 3.0% m/v is diluted to 950 mL. What is its final mass/volume percent?

Answers

1. Solubility is the amount of a solute that can dissolve in a given amount of solute, typically 100 mL. The solubility of solutes varies widely.

3. 26.5%

5. 0.203%

7. 52.4 mL

9. 0.00321 g

11. 0.130 M

13. 0.208 M

15. 37.6 g

17. 0.496 L

19. 0.262 mol; 37.5 g

21. 1.13 M C2H5OH; 1.13 mol of CO2; 49.7 g of CO2; 27.7 L of CO2

23. 1.0 Eq/L

25. 0.200 M