3.3 Solution’s Different Concentrations

Introducing solutions

Solutions are homogeneous (single-phase) mixtures of two or more components. For convenience, we often refer to the majority component as the solvent; minority components are solutes. But there is really no fundamental distinction between them.

Solutions play a very important role in Chemistry because they allow intimate and varied encounters between molecules of different kinds, a condition that is essential for rapid chemical reactions to occur. Several more explicit reasons can be cited for devoting a significant amount of effort to the subject of solutions:

• For the reason stated above, most chemical reactions that are carried out in the laboratory and in industry, and that occur in living organisms, take place in solution.
• Solutions are so common; very few pure substances are found in nature.
• Solutions provide a convenient and accurate means of introducing known small amounts of a substance to a reaction system. Advantage is taken of this in the process of titration, for example.
• The physical properties of solutions are sensitively influenced by the balance between the intermolecular forces of like and unlike (solvent and solute) molecules. The physical properties of solutions thus serve as useful experimental probes of these intermolecular forces.

We usually think of a solution as a liquid made by adding a gas, a solid or another liquid solute in a liquid solvent. Actually, solutions can exist as gases and solids as well. Gaseous mixtures don’t require any special consideration. Solid solutions are very common; most natural minerals and many metallic alloys are solid solutions.

Still, it is liquid solutions that we most frequently encounter and must deal with. Experience has taught us that sugar and salt dissolve readily in water, but that oil and water don’t mix. Actually, this is not strictly correct, since all substances have at least a slight tendency to dissolve in each other. This raises two important and related questions: why do solutions tend to form in the first place, and what factors limit their mutual solubilities? These questions will be answered in Chem II course but here we will discuss the concentration and reactions of aqueous solutions.

2  How solution concentrations are expressed

Concentration is a general term that expresses the quantity of solute contained in a given amount of solution. Various ways of expressing concentration are in use; the choice is usually a matter of convenience in a particular application. You should become familiar with all of them.

The concentration of a solution is referred to as the ratio of the solute amount to the amount of the solution (or a solvent in some specific cases).

The below is the table summarizing all types of concentrations formulas used in this chapter

[mol/L]

Also can be expressed as M or molar

(Molality)

= [moles of the solute]/[Kilogram of solvent

Moles (mol)

Kilogram (kg)

[mol/kg]

Also can be expressed as m or molal

Equivalent molarity of the electrolytes

= [milliequivalents/Liter of solution]

Milliequivalents (mEq)

Liter (L)

[mEq/L]

Dilution

C₁xV₁ = C₂xV₂ or sometimes is given in:

M₁xV₁=M₂xV₂

Where C₁ and M₁ are the concentration and the molarity of the stock solution respectively. C₂ and M₂are the concentration and molarity of the diluted solution respectively

V₁ is the volume of the stock solution and V₂ is the volume of diluted solution

Moles (mol)

Liter (L)

[mol/L]

Also can be expressed as M or molar

Part per million (ppm)

Part per billion (ppb)

ppm = {[mass of solute] / [mass of solution]} x 1 x 10⁶

ppb = {[mass of solute] / [mass of solution]} x 1 x 10⁹

ppm

ppb

Let us now go over some workout problems in some details covering all the formulas.

1. (Mass/volume)%
2. What is the mass/volume percent by of a solution formed by mixing 25.0 grams of sodium chloride NaCl with 455 mL of water?

Solution:

(Mass/volume)% = {[mass of the solute]/[volume of the solution]} x100%

                               = {[25.0 g NaCl] / [455 mL solution]} x100% = 5.494505494505495% (answer from calculator and it should be rounded off to the correct significant figures which is 3.

Final answer after the significant figures rule consideration is 5.50%  (g/mL %)

Calculate amount of grams of solute potassium nitrate aqueous solution which is 38.5% (mass/mL%) if the solution made with 1.00 Liter of water.

Solution:

Let us set up the formula and replacing it with the given data:

(Mass/volume)% = {[mass of the solute]/[volume of the solution]} x100%

First the volume should be converted from Liter into milli Liter (mL)

[38.5 g/mL]% = { [X] / [1000 mL] } x 100%

Cross x Multiply

[38.5 g/mL] / [100]   =  [X] / [1000 mL]

   X  = 385 g potassium nitrate KNO₃

Verifying the answer:

(Mass/volume)% = {[mass of the solute]/[volume of the solution]} x100%

                               = [385 g / 1000 mL] x100% = (38.5 g/mL)%

Conclusion the answer is correct.

2. (Mass/mass)% or mass%

A solution is formed by adding 48 g of Lithium sulfate to 250 g of water. What is the percent by mass of Lithium sulfate?

         (Mass/mass)% or mass% = [mass of the solute]/[mass of the solution]x100%

Note that:

Mass of solution = mass of solute + mass of solvent

(Mass/mass)%  = {[48 g of Lithium sulfate] /[48 g of Lithium sulfate + 250 g water]} x 100%

                         = 16.10738255033557% from calculator and it should be rounded off to the correct significant figures

Final answer = 16% [g/g]%

What is the mass of solvent (water) in grams for a solution made of 10.5% mass percent of each component in the mixture formed by adding 2.00 g of Aluminum nitrate in water?

         (Mass/mass)% or mass% =[mass of the solute]/[mass of the solution]x100%

Note that:

Mass of solution = mass of solute + mass of solvent

Mass of the whole solution is not known and it will be referred to as X

[10.5 / 100] = [2.00 g Aluminum nitrate] / [ X ]

Cross / Multiply:

[0.105] * [ X ] = 2.00

X = [2.00 /0.105] = 19.04761904761905 = 19.0

X = amount of water in grams = 19.0 – 2.00 = 17.0 grams water

Verifying the answer:

 [Mass/mass]%  = [2.00 /(2.00 + 17.0)] x 100%  = 10.52631578947368%  = 10.5%

The answer is correct.

3. (Volume/volume)% or volume%

What is the percent by volume of a solution formed by added 25 L of ethanol to 155 L of water?

(Volume/volume)% or volume% = [volume of the solute]/[volume of the solution]x100%

Note that:

Volume of the solution = volume in solute + volume of the solvent

(Volume/volume)% = [ 25 L ] /[ 25L + 155 L ] x 100 % = [ 25 L ] / [ 180 L ] = 13.88888888888 %. The answer is from calculator and it should be rounded off to the correct significant figures

Final answer is 14 % (two significant figures).

2. A 5.50% volume percent solution is made by adding certain amount of volume of a solute acetone to 255 mL water as solvent. Calculate the amount of the acetone in mL added.

(Volume/volume)% or volume% = [volume of the solute]/[volume of the solution]x100%

Note that:

Volume of the solution = volume in solute + volume of the solvent

Amount of the solute is unknown and it will be referred to as X

5.50%   = [ X ] / [ X + 255 mL ]

0.0550 = [ X ] / [ X + 255 mL ]

0.0550 [ X + 255 mL ] = [ X ]

0.0550 X + (0.0550*255) = [ X ]

Solving for X

0.0550*255 = X – 0.0550 X = 0.945 X

X = 14.8 mL (3 significant figures)

Verifying the answer:

Volume/volume)% = { [ 14.8 mL ] / [ 14.8 mL + 255 mL] } x 100% = 5.485544848035582 % rounded to 3 significant figures = 5.50 %. The answer is correct.

4. Molarity

Molarity = [moles of the solute]/[liter of solution]

1. Calculate the molarity of solution made by dissolving 35.0 grams of sodium chloride solid in 97890. mL distilled water.

Molarity = [moles of the solute]/[liter of solution]

First one has to calculate the number of moles of sodium chloride with given mass of 35.0 grams. Therefore, one has to use the molar mass of sodium chloride in calculating number of moles with given number of grams.

Molar mass of sodium chloride = NaCl using the atomic masses of Na and Cl, the molar mass is:

23.0 g/mole + 35.5 g/mol = 58.5 g/mol

Number of moles of sodium chloride = (35.0 g) / (58.5 g/mol)  = 0.598 moles NaCl

Second one has to convert the mL into Liters by dividing by 1000 =(97890. mL) / (1L / 1000 mL) = 97.890 L

Molarity = [0.598 moles NaCl] / [97.890 L} = 0.006108897742363877 mol/L. This the calculator’s answer and it will be rounded off to the correct significant figures ( 3 significant figures).

Molarity = 0.00611 mol/L or 0.00611 M or 0.00611 molar.

2. A solution of a molarity of 2.675 M is prepared by dissolving 8.50 gram of magnesium nitrate in certain amount of distilled water as a solvent. Calculate the amount of the distilled water.

        First one has to calculate the number of moles of magnesium nitrate with given mass of 8.50 grams.

Therefore, one has to use the molar mass of magnesium nitrate in calculating number of moles with given number of grams.

Molar mass of magnesium nitrate = Mg(NO₃)₂ using the atomic masses of Mg and 2 (NO₃), the molar mass is: 24.3 g/mol + 2x (14.0 g/mol) + 2 (3 x 16.0 g/mol) =148.3 g/mol

Number of moles of magnesium nitrate = [8.50 g Mg(NO₃)₂] / [148.3 g Mg(NO₃)₂ / mol Mg(NO₃)₂]

Number of moles of magnesium nitrate = 0.0573 moles

2.675 M = 2.675 mol /L = [0.0573 mol] / [X]

X = amount of the solution or solvent in Liters

Solving for X, one can obtain:

2.675 X = 0.0573

X = 0.0573 / 2.675 = 0.02142056074766355 Liters. The correct rounded off answer has to have three significant figures. The final correct answer is 0.0214 Liters

Verifying the answer: = [0.0573 mol] / [0.0214 Liters] = 2.6775700934579439 = 2.678 mol/L

The answer is correct. Note that the verified answer is not exactly the given molarity of 2.675 mol/L. This is because of the in between rounding off and rounding off the atomic masses.

3. 10.00 x 10²⁶ molecules of acetone dissolved to make 3500. mL of water of solvent. Calculate the molarity of acetone solution.

First one has to calculate the number of moles of acetone with given number of molecules.

Number of moles of acetone = [10.00 x 10²⁶ molecules of acetone] x [1 mole acetone/6.022 x 10²³ molecules of acetone]

One has to use the Avogadro’s number of 6.022 x 10²³ to calculate number of moles:

1 mole of acetone = 6.022 x 10²³ molecules of acetone

Moles of acetone = 1660.577881102624 moles

The volume of solution is 3500. mL = 3.500 L (volume is converted into Liters)

Molarity = [1660.577881102624 moles] / [3.500 L] = 474.4508231721783 mol/L = 474.5 M or 474.5 mol/L or 474.5 molar.

The answer is rounded to the correct 4 significant figures.

5. Molality =  [moles of the solute]/[Kilogram of solvent

1. What is the molality of a solution made of 9.384 grams of aluminum nitrate has been dissolved in 5500.0 g water

First the number of moles of aluminum nitrate is calculated using the molar mass of aluminum nitrate.

Molar mass of aluminum nitrate= Al(NO₃)₃ = Al + 3N + 9O = 27.0 g/mol + (3 x 14.0) g/mol + (9 x 16.0 g/mol) = 213.0 g/mol

Number of moles = [9.384 grams of aluminum nitrate] / [mol aluminum nitrate /(= 213.0 g/mol)] = 0.04405633802816902 moles aluminum nitrate.

Second to convert 5500.0 g water into kilo grams water by diving by 1000 = 5.500 kg water

Molality = [0.04405633802816902 moles aluminum nitrate] / [5.500 kg water] = 0.008010243277848913 mol/kg. The answer should be rounded off to the correct significant figures of 4.

The final correct answer is 0.008010 mol/kg or m or molal.

2. What mass of water is required to dissolve 45.0 grams calcium chloride to prepare a 2.50 m solution?

First the number of moles of calcium chloride is calculated using the molar mass of calcium chloride.

Molar mass of calcium chloride CaCl₂ = Ca + 2 Cl = 40.0 g/mol + 2x 35.5 g/mol = 111.0 g/mol

Number of moles CaCl₂ = [45.0 g calcium chloride] / [111.0 g/mol] = 0.4054054054054054 moles = 0.405 moles CaCl₂

Second the molality formula is used:

2.50 m = [2.50 CaCl2mol / kg water] = [0.405 moles CaCl₂] / X

Where X is the amount of water

Solving for X yields:

X = 0.405 / 2.50 = 0.162 kg water

Verifying:

Molality = [0.405 moles CaCl₂] / [0.162 kg water] = 2.50 m

6. Equivalent molarity of the electrolytes

Milliequivalents (mEq) calculations are covered in greater details in the link below:

https://www.researchgate.net/file.PostFileLoader.html?id=546acea8d4c1182b638b473a&assetKey=AS:273639646138393@1442252184176

Milliequivalents calculations cover the three topics:

1. Converting milliequivalents to weight
2. Converting weight to milliequivalents
3. Converting milligrams% into milliequivalents

Each electrolyte  such as Na⁺ is measured in equivalent (Eq).For example 1 mole of an ionic compound of NaCl dissolved in water will produce 1 mole of Na⁺ 1 mole of and Cl^–. Each Na⁺ and Cl^– which equal to 1 equivalent each.

The table below illustrates the relationship between the electrolytes charge and number of the equivalents.

1.  Converting milliequivalents to weight

What is the concentration of a solution contains 8.50 mEq/L of NaCl?

First the molar mass of NaCl should be calculated. NaCl molar mass is calculated in the above examples and it equals 58.5 g/mol

Second use the general formula for the calculation of the milligram/L from mEq/L:

Mg/L = [(mEq/L) x (atomic mass or molar mass)] / [number of the equivalent)

Mg/L = [(8.50 mEq/L) x (58.5 g /mol)] / [1 Eq/ 1 mol] = 497.25 mg / L

Final answer is 497 mg / L (3 significant figures).

2. Converting weight to milliequivalents

How many mEq of NaCl are in 10.5 g of NaCl?

First the molar mass of NaCl should be calculated. NaCl molar mass is calculated in the above examples and it equals 58.5 g/mol

Second use the setup that relates mEq to mass of electrolytes using molar mass:

mEq = {[10.5 g of NaCl] / [58.5 g NaCl/mol]} * [Eq /1 mol] * [1000 mEq/ 1Eq] = 179 mEq

3. Converting milligrams% into milliequivalents

Convert the expression 15.0 mg% of Ba²⁺ to mEq/L

Ba²⁺ atomic mass = 137.3 g/mol

        Ba has 2+ ion and hence 2 Eq

        Equivalent mass of Ba²⁺ = [137.3 g/mol] / [2 Eq/mol] = 68.65 g/Eq = 68.7 g per Eq

        1mEq Ba²⁺ = [68.7 g] / [1 g/1000 mg] = 0.0687 mg

        15.0 mg% = [15.0 mg Ba²⁺] / [100 mL solvent] = [15.0 mg/ 100 mL]*[1000 mL/1L] = 150. mg/L

0.0687 mg/ 1 mEq  =  150. mg/ X mEq

Solving for X = 150. / 0.0687 = 2183.406113537118 mEq/L rounded off to 2180 mEq/L (3 significant figures) or expressed in scientific notation 2.18 X 10³ mEq/L

7. Dilution

C₁xV₁ = C₂xV₂ or sometimes is given in:

M₁xV₁=M₂xV₂

Where C₁ and M₁ are the concentration and the molarity of the stock solution respectively. C₂ and M₂are the concentration and molarity of the diluted solution respectively

V₁ is the volume of the stock solution and V₂ is the volume of diluted solution

The stock solution is considered as the most concentrated solution.

 Examples:

1. If 75.5 mL of water are added to 875.0 mL of a 0.125 M HNO₃ solution, what will be the new molarity of the diluted solution?

C₁xV₁ = C₂xV₂

Total volume = V₂ = 75.5 + 875.0 = 950.5 mL

0.125 M x 875.0 mL = C₂ x 950.5 mL

C₂ = [0.125 M x 875.0 mL] / [950.5 mL] = 0.115 M

2. If 500.0 mL are added to 0.150 M HCl solution until the final volume is 650.0 mL, what will be the molarity of the diluted solution?

C₁xV₁ = C₂xV₂

Total volume = V₂ = 650.0 mL

0.150 M x 500.0 mL = C₂ x 650.0 mL

C₂ = [0. 0.150 M x 500.0 mL] / [650.0 mL] = 0.115 M

3. How much water should be added to 985 mL of a 6.55 M NaOH solution to make a 1.00 M solution?

C₁xV₁ = C₂xV₂

Total volume = V₂ = X + 985 mL

X = volume of water

6.55 M x 985 mL = 1.00 M x V₂

V₂ = [6.55 M x 985 mL] / [1.00 M] = 6451.75 mL

X + 985 mL = 6451.75 mL

X = 5466.75 mL = 5470 mL rounded off to 3 significant figures

4. 450.5 mL of a 2.660 M KNO₃ solution. If one boils the water until the volume of the solution is half of its amount, what will be the molarity of the solution after the boiling?

C₁xV₁ = C₂xV₂

Total volume = V₂ = 450.5/2 mL = 225.3 mL

450.5 mL x 2.660 M KNO₃ = C₂ x 225.3 mL

C₂ = [450.5 x 2.660] / [225.3] = 5.319 M [Boiling the stock solution had led to higher concentration of the new boiled solution]

After going over couple examples of the concentrations, let us look at some available simulation activities for the concentrations of the solution.

A Phet molarity simulation activity is given below:

In this simulation, the students will use the information below:

Solution volume is the combined volume of solute and water.

By design, not all solutions will reach saturation. The number of moles that can be added is limited to the range of 0.2-1.0 moles so that students can explore some solutions for the full concentration range (0-5 M).

Drink mix is assume to have the same solubility as sucrose.

Solubility of each solution listed was calculated at 25⁰ C, except for AuCl₃ and Drink mix, which were based on data taken at 20⁰ C.

Activity:

Determine the qualitative relationships between molarity, moles, and liters before completing quantitative problems or data collection.

The simulation demonstrates saturation but does not explain why different solutes have different solubilities. The Drink Mix example provides a real-world link to the concept of concentration to help the  students make connections to the chemical examples. Determine the saturation molarity in mol/L for all salts (solutes) in the simulation.

Questions:

1. Which one of the salts (solutes) used has the largest saturation concentration

2. Which one of the salts (solutes) used has the smallest saturation concentration

3. Arrange the salts (solutes) in increasing order of saturation concentration. Explain the difference in the saturation concentrations among the salts (solutes).

Several You Tube videos discussing the concentration of solution:

(Mass) %

(Mass/volume) %:

             (Volume/volume) %

Molarity:

Molality:

https://www.youtube.com/watch?v=uj1u7Nx9JUc

https://www.youtube.com/watch?v=uj1u7Nx9JUc

Dilution:

Part Per Million (ppm) and Part Per Billion (ppb):

Parts per million (ppm) and parts per billion (ppb) may be converted from one to the other using this relationship: 1 part per million = 1,000 parts per billion.

ppm = {[mass of solute] / [mass of solution]} x 1 x 10⁶

ppb = {[mass of solute] / [mass of solution]} x 1 x 10⁹

Examples:

The solubility of NaCl is 284 grams/100 grams of water.  What is this concentration in ppm?

                                    284 X 1,000,000 = 2840000 ppm

                                    100

The solubility of AgCl is 0.008 grams/100 grams of water.  What is this concentration in ppm?

.008 X 1000000 = 80 ppm

                                    100

A certain pesticide has a toxic solubility of 5.0 grams/Kg of body weight.  What is this solubility in ppm?

                                    1 Kg X 1000 g = 1000 g         __5__ X 1,000,000 = 5000 ppm

                                       1         1Kg                           1000

Change 50 ppm to ppb.

            50 ppm = 50 X 1,000,000 = 50,000,000        50,000,000 X 1,000=50,000,000,000

X(ppm) = X(ppb) / 100

ppb = 1000 ppm                                      

50 ppm = 50 x 1000 ppb = 50,000 ppb

How many parts per million (ppm) is 1mg/L?

1. ppm