UNIT 4 – Solutions · Web viewChemistry 30S Solutions Unit Four Describe and give examples of various types of solutions. Describe the structure of water in terms of electronegativity

Chemistry 30S

SolutionsUnit Four

Describe and give examples of various types of solutions. Describe the structure of water in terms of electronegativity and the polarity of

its chemical bonds. Explain the solution process of simple ionic and covalent compounds, using

visual, particulate representations and chemical equations. Explain heat of solution with reference to specific applications. Perform a lab to illustrate the formation of solutions in terms of the polar and

nonpolar nature of substances. Differentiate among saturated, unsaturated, and supersaturated solutions. Construct, from experimental data, a solubility curve of a pure substance in

water and use a graph of solubility data to solve problems. Explain how a change in temperature affects the solubility of gases. Explain how a change in pressure affects the solubility of gases. Perform a lab to demonstrate freezing-point depression and boiling-point

elevation, and explain freezing-point depression and boiling-point elevation at the molecular level.

Differentiate among, and give examples of, the use of various representations of concentration. Solve problems involving calculation for concentration, moles, mass, and

volume. Prepare a solution, given the amount of solute (in grams) and the volume of

solution (in millilitres), and determine the concentration in moles/litre. Solve problems involving the dilution of solutions. Perform a dilution from a solution of known concentration.

Gr. 11 Chemistry Page 2 Solutions

Solutions

Classification of MatterRecall from the last unit:

Matter

Pure Substances(constant composition)

Mixtures(variable composition)

Elements Compounds Homogeneous(same properties)

Heterogeneous(different properties)

__________________________________________________________________________________. The particles in a heterogeneous mixture are not evenly distributed and individual particle are often distinguishable (e.g. __________________________). The particles of a homogeneous mixture are evenly distributed and cannot be easily distinguishable (e.g. _________________________________).

Heterogeneous Mixtures

Heterogeneous mixtures will _______________________ if left to stand. There are two main types of heterogeneous mixtures:

1. Suspensions o Mixture of _________________________________ that will settle out upon standing.o Components can be separated by a _______________________.o Example: _______________________________________________________________

2. Colloids o Mixture of ___________________________________.o Have a ________________________________ appearance (can’t see through)o Particles are ________________________________ throughout a medium.o Examples: _______________________________________________________________


Solutions

Homogeneous Mixtures

Homogeneous means to be _______________________ throughout a sample of material, such that every part of the material is exactly the same as any other part in terms of ____________________________________________.

Solution: ____________________________________________________________________________________________________________________________________________________

Solutions are made of two components:1. Solvent : is the __________________________________, and is usually the substance present

in the ______________________________________ in a solution.

2. Solute : is the ___________________________________, and is usually the substance present in ________________________________________ in a solution.

Example of a solution: Consider making lemonade by mixing lemonade crystals into a glass of water. ________________________________________________________________________________________________________________________________

5 Characteristics of a Solution

1. Solutions are ________________________ – they look the same throughout (same properties).

2. Solutions are ________________ – the solute will not settle out over time.

3. Both the solute and solvent pass through a ___________________ when filtered.

4. Solvents and solutes may be ______________________________________________.

5. Solutions are considered to be a single phase even though the components may have been in different phases before the solution was formed.


Types of Solutions

There are 3 common physical states of matter (solid, liquid, and gas), therefore, there are 9 possible combinations of solvent-solute pairs.

Solute Solvent Example of Solution

Solid Liquid________________________________________________________________________________________________________________________

Liquid Liquid ____________________________________ (gas line antifreeze)Ethylene glycol in water (engine antifreeze)

Gas Liquid __________________________________________________Dissolved oxygen in water (supporting aquatic life)

Solid Solid

__________________________ (brass)14 karat gold (14 partsgold and 10 parts other metal)______________________ (sterling silver)____________ (iron and carbon)

Liquid Solid ____________________________________ (dental amalgams)

Gas Solid ______________________________________________Pumice stone

Solid Gas

________________________________________Any substance that is normally solid at room temperature and sublimes________________________________________

Liquid Gas ________________________________________________________________________________________

Gas Gas ____________________________________________

Note: A substance is considered __________________ if it will dissolve in a specific solvent.A substance is considered ______________________ if it does not dissolve in a specific solvent.


___________

Practice: Types of Solutions

1. Kool-Aid _______________________________________

2. Tooth fillings ___________________________________

3. Pepsi _________________________________________

4. Cigarette Smoke ________________________________

5. Rubbing Alcohol ________________________________

6. Marshmallow __________________________________

7. Air ___________________________________________

8. Sterling silver ring ______________________________

9. Humidity _____________________________________


The Solvation Process and Heat of Solution

Solvation occurs when _________________________________________________. When a solute dissolves in a solvent, the individual particles of the solute separate from the other particles of the solute. They then move between the spaces of the solvent particles. The solvent particles must collide with the solute particles and forces of attraction between solute and solvent particles “hold” the solute particles in the spaces.

Heat of Solution

When attractive forces are _______________________, energy is ____________________. Therefore, the separation of solute particles from one another and the separation of solvent particles from one another are both _______________________________ processes (require energy).

The attraction between solute and solvent particles during the solvation process is _______________ (release energy).

Whether energy is absorbed or released in the overall net process of solution formation depends on the balance between these two processes.

The net energy change is called the __________________________________.

ENDOTHERMIC:If the amount of energy absorbed is greater than the amount of energy released, then the overall solution becomes endothermic.

Example: NH4NO3(s) + heat NH4+

(aq) + NO3-(aq)

Example: _______________________________________

EXOTHERMIC:If the amount of energy absorbed is less than the amount of energy released, then the overall solution becomes exothermic.

Example: CaCl2(s) Ca2+(aq) + 2Cl-

(aq) + heat

Example: _______________________________________


Solvation Process of Ionic CompoundsIonic compounds consist of charged ions held together by electrostatic forces of attraction (________________________________ bonds). They are generally _________________, crystalline structures at ______________________________.

Example: Sodium chloride (salt) being dissolved in water.

NaCl (salt) is the __________________

Water is the _____________________

When a solute is placed into a solvent, the solvent particles completely surround the surface of the solute particles. Polar water molecules orientate themselves around each exposed ion on the crystal surface (lattice) of the solid. The positive end of the water molecule orientates itself toward the negative chloride ion Cl- and the negative end of the water molecule toward the positive sodium ion Na+.


The Solvation Process

If the solute is soluble, the attraction between the solvent molecules and the solute ions gradually increases to the point where it finally exceeds the forces holding the ions into the crystal lattice. As a result, the solute ions become completely surrounded by the solvent molecules.

There are three steps to the dissolving process:

1. The solvent particles must _______________________ to make room for the solute particles. This process ________________________ to overcome the forces of attraction between solvent particles and is therefore endothermic.

2. The solute particles must ______________________from the other solute particles. This process ______________________ to overcome the forces of attraction between the solute particles and therefore is also endothermic.

3. When the solute particles move between the solvent particles the forces of attraction between the solute and solvent take hold and particles “_________________” back and move closer. This process ______________________________ and is therefore exothermic.

The separation of ions is called _________________________, whereas the process of surrounding the solute particles with solvent particles is called _______________________________.

If the solvent is water, this process is called ___________________________. The solute particles are said to be hydrated.


Solvation Process of Covalent CompoundsWhen covalent compounds are placed in a solvent, the entire molecule is pulled away from the solid structure.

Aqueous Solutions and Dissociation

Aqueous Solutions

Solutions in which water is the solvent are called __________________________________. They are common because water can dissolve many substances. Water is often called the ____________________________________ for this reason.

Dissociation

The separation of _____________ and ________________ ions is called ______________________.

NaCl (s) H2O(l)→

Na+(aq)+Cl-

(aq)

When ionic compounds like NaCl dissolve in water, they no longer exist as a ____________________, but as freely moving _________________ that are generally not associated with each other.

Molecular compounds like sugar do not separate into charged particles, but dissolve as whole molecules, and therefore do not dissociate.

C11H22O11(s) C11H22O11(aq)

Dissociation may occur when ionic compounds are:1. In the molten state (melted at EXTEMELY high temperatures no water used to dissolve). 2. In the aqueous state.

a. The sodium and chlorine ions dissociate as the water molecules surround them.b. A crystal of an ionic compound will not dissolve if the attractive forces within the crystal

are stronger than the attractive forces of the water.


Dissociation EquationsExample:Write the equation for dissolving solid magnesium chloride, MgCl2, in water.

Step 1: Determine if the compound is ionic or molecular

Step 2: Write the ions that will be formed in the dissociation.

Step 3: Use the subscripts from the formula to indicate the coefficient that will balance the charges.

Step 4: Write the equation using the appropriate state for the compound, (s) or (aq), for each dissociated ion.

Examples:

Write chemical equations for the dissolving and dissociation of the following compounds in water.

1. Solid aluminum sulphate, Al2(SO4)3

2. Liquid methanol, CH3OH

3. Solid copper(II) chromate (CuCrO4)

4. Solid ammonium chloride (NH4Cl)


Dissociation Equations Worksheet

1. What type of compound will dissociate in water? What type will not?

2. What part of the water molecule would be attracted to a solute ion such as Cl-? Na+?

3. Write the equation for the dissolving of each of the following in water:

a. PbSO4(s)

b. C6H12O6(s) _____ __

c. KBr(s)

d. NaF(s)

e. CH3OH(l)

f. calcium chloride (s)

g. sodium carbonate (s)

4. Describe what occurs during the process of an ionic solid dissolving into water. Include what happens to the solute and to the solvent.


Polar vs. Non-Polar

Miscibility_____________________ is the property of mutual solubility of two liquids. Two liquids dissolve within one another to make a solution.

Water and antifreeze are completely __________________ (they mix together completely). Ethyl ether and water dissolve in each other to some extent and are called partially miscible. Water and oil do not dissolve in each other and are called __________________ because they

separate into layers when standing. Some metals (gold and silver) are mutually miscible and form solid-solid solutions; these metal

solutions are called ______________________.

Solvent-Solute Combinations Based on Polarity

There are 4 possible solvent-solute combinations based on the polarity of the solvent and solute particles.

1. ______________________________________________o Example: Dissolving NaCl in H2Oo The polar solvent particles solvate the polar solute particles by attaching themselves due to

the polar attractions (attraction between charged ions of the salt and partial charges of the atoms in the water molecule).

2. ______________________________________________o Example: Dissolving oil in H2Oo Since the solute particles are non-polar, the solvent particles are not attracted to them – they

are repelled. Forces of attraction between the water molecules are greater than to the non-polar substance.

o Solvation to any extent in unlikely.

3. ______________________________________________o Example: Dissolving NaCl in oil.o Since the solvent particles are non-polar they are not attracted to the polar solute particles –

they are repelled.o Solvation to any extent is unlikely.

4. ______________________________________________o Example: Dissolving wax in oil.o A solution is formed because there is no repulsion between the non-polar solute and non-

polar solvent.o The degree of solvation will not be as much as between a polar solute and solvent.


Solvation Rate

Solvation rate is affected by two factors:

1. The amount of _________________________________ exposed to fresh solvent.o Increasing the exposed surface area _________________________ the rate of solvation.o Surface area (SA) can be increased by:

a. Breaking the crystal into very ______________________ thereby increasing the SA.b. __________________________ the mixture as the solute dissolves.

This moves the solvent saturated with solute away from the surface of the solid so that fresh solute can come into contact with the solid.

Example: sugar cube vs. granulated sugar.

2. ______________________________________________ of both the solute and solvent particles.o _________________________ is a measure of the average kinetic energy of the particles.o The _____________________ the solvent particles are moving, the more rapidly they

circulate and move away from the solid material thereby exposing more fresh surfaces.o Increasing the kinetic energy _____________________________ of the molecules making

it easier for particles to be removed from the crystal.

Example: sugar cube in hot water vs. sugar cube in cold water


Solubility

Solubility: _________________________________________________________________________ _________________________________________________________________________

There are three ways in which we can describe how soluble a substance is in a solution:

Saturated Solutiono A solution that holds the _____________________________ of solute possible at a given

temperature. o If you add more solute to the solution, it will ______________________. The extra solute will fall

to the bottom of the container. o For example, if you add more sugar to a saturated solution the added sugar will simply drop to the

bottom of the container (it will _________________________).

Unsaturated Solutiono A solution that holds ___________________________ the maximum amount of solute possible at

a given temperature. o This solution can _____________________________________. o If additional solute is added to the solvent, it will _______________ until the solution is saturated.o For example, if you add more sugar to an unsaturated sugar solution, the sugar will continue to

dissolve until the solution reaches its saturation point.

Supersaturated Solutiono A solution that holds ________________________ the maximum amount of solute possible at a

given temperature. o The solution is very ____________________, and if you add just one crystal of solid, much more

solute will precipitate out of the solution. You can prepare a supersaturated solution by first making a saturated solution with excess

solute on the bottom of the container, then slowly heating the solution until all the solute is dissolved.

When all the solute is dissolved, begin slowly cooling the solution. When the solution reaches room temperature, there will be more dissolved than was

initially dissolved when the solution was saturated. Add a single crystal of solute or tap the side of the container and watch the excess solute

precipitate out of solution

Demonstrations of Supersaturated solutions:

http://www.youtube.com/watch?v=1y3bKIOkcmk

http://www.youtube.com/watch?v=6SZaEzmDkYo&feature=related


http://www.youtube.com/watch?v=6SZaEzmDkYo&feature=related

http://www.youtube.com/watch?v=1y3bKIOkcmk

Temperature and Solubility of Gases

Consider a glass of Pepsi on the counter (T = 22˚C) and a glass of Pepsi in the fridge (T = 5˚C). CO2 gas (bubbles) are escaping much more rapidly from the glass on the counter than the glass in the fridge. Since the CO2 is escaping from the solution (Pepsi) into the atmosphere, the quantity of CO2 in the solution is decreasing; therefore the solubility is decreasing in both glasses.

However, the CO2 is escaping more rapidly from the higher temperature glass (counter) indicating that the solubility of the gas is lower when the solution is at a higher temperature. These observations can be summarized in the following relationship:


This relationship is important to fish as they depend on dissolved oxygen to maintain cellular respiration. If the water temperature increases, the solubility of the dissolved oxygen gas decreases and fish may die (suffocate). This is a concern of thermal pollution from power plants.

Pressure and Solubility of Gases

The pressure above a liquid affects the solubility of gasses in liquid. When you open a pop bottle, there is a sudden rush of gas from the bottle. When the bottle is opened, the built up pressure is released and the solubility of the carbon dioxide is released.

Increasing pressure increases the solubility of a gas in a liquid because increasing the pressure forces the gas particles into contact with the liquid. As the gas particles contact the liquid, forces of attraction from the liquid cause the gas to condense and dissolve.

The effect of pressure on solubility is important for SCUBA divers. Underwater, for every 10 meters of depth the pressure increases by 1 atmosphere. Eventually, the pressure becomes so great that even the slightly soluble nitrogen gas dissolves in the diver’s blood. If the diver comes up too quickly (decreasing pressure causing a decrease in the solubility of the dissolved gases in blood), the nitrogen will begin to come out of solution and form bubbles in the blood vessels. Gas bubbles become trapped in capillaries and block the flow of blood. This is known as decompression sickness.


Consider a can/bottle of Pepsi:

1. Is this a solution? _______________

2. What is (are) the solute(s)? _________________________________________________________

3. What is the solvent? ______________________________________________________________

4. What type of solution is this? _______________________________________________________

5. Are the solutes and solvent polar or non-polar? _________________________________________

6. Why does the drink make a popping or fizzing noise when you open it? ______________________

________________________________________________________________________________

Carbonated beverages contain CO2 gas dissolved in an aqueous solution (fizz). In bottling the beverage, CO2 is dissolved in the solution at a pressure higher than atmospheric pressure. When the cap is opened, the decrease in pressure above the solution results in decrease solubility of the CO2, therefore CO2 escapes.

Note: The solubility of solids and liquids are not affected by a change in pressure but the solubility of a gas in a liquid is greatly affected by pressure changes.


Solubility Curves

The solubility of most substances _________________________ as the temperature of the solvent _________________________. For some substances, the reverse occurs.

Consider the graph below, the solubility of all compounds, except for Ce2(SO4)3, increases as the temperature increases.

Solubility curves graph the _____________________________________________________ of a solid dissolved in a liquid solvent, water. A curve of best fit is drawn through the points.

From a solubility curve, we can predict the solubility of a solution at a specific temperature, or the temperature at which a solution will have a specific solubility. That is, we can ____________________ points on the curve to find the information we need.


Solubility UnitsSolubility is the ____________________________________________________________________ (or the amount of solute needed to make a saturated solution, under certain conditions).

The units of solubility are usually in terms of ____________________________________ (in grams), per ____________________________. Since the density of water is 1.00 g/mL, the solubility in water is sometimes given as grams of solute per 100 mL of water. Another set of units used for solubility is grams of solute per Litre of solvent, for a liquid solvent.

The solubility of a solute must be determined experimentally. One method used to determine the solubility of a substance at a specific temperature is to add a known mass of solute to a known mass, or volume, of solvent. The mixture is heated until all the solute dissolves. When the solute has dissolved, the solution is cooled until the first signs of solid crystals appear. The temperature at this point is recorded and the solubility is converted to the desired solubility units.

EXAMPLE 1:

If 25.0 g of a solute is the maximum amount of solute that can dissolve in 40.0 g of solvent at a certain temperature, what is the solubility in grams of solute per 100 g of solvent?

Solution:

You can solve this problem by using ratios. If you solve this problem using ratios, it would look like this:

25 g of solute = mass of solute 40 g of solvent 100 g of solvent

For this type of problem, it is best to use the following:

Mass of solute x mass of solvent needed Mass of solvent

To solve this problem, the equation above is a modified form of a ratio.

25 g of solute x 100 = 62.5 g of solute / 100 g of solvent40 g of solvent

You may choose the method that you like best.


Solubility Units

EXAMPLE 2:

If 30.1 g of a solute can dissolve can dissolve in 350.0 mL of water at a certain temperature, what is the solubility of the substance in g/100g water?

EXAMPLE 3:

If 39 grams of solute is the maximum amount of solute that can dissolve in 55 grams of water at a certain temperature, what is the solubility in g/100g of water?


Solubility CurvesSolubility is _____________________________________. We can find the solubility of a substance at various temperatures and plot this data on a graph. The result is what is known as a solubility curve. The diagram below shows an example of a solubility curve for potassium nitrate.

Each point _____________________________ represents a saturated solution of potassium nitrate.

The area _______________________________________ represents quantities that produce an unsaturated solution at that temperature.

The points _____________________________ indicate either a supersaturated solution or a saturated solution with some remaining undissolved solute.


Example: Solubility Curves Problems

Answer the following questions using the diagram below.

1. What is the solubility of Ce2(SO4)3 at 25C?

2. What temperature is the solubility of NaCl approximately 35 grams/100 grams of H2O?

3. I have a solution of KCl with a solubility of 42 grams per 100 grams of H2O at 50C. Is my solution unsaturated, saturated or supersaturated? How do you know?

4. I dissolved 20 grams of NaNO3 in 25 grams of water. What temperature was my solution at?

*To solve this type of problem, you need to know the mass of the solute for 100g of water. Use the following ratio to help solve the problem*


Solubility CurvesWorksheet

1. What is the solubility of potassium nitrate, KNO3, at 44C?

2. 25 g of potassium nitrate is dissolved in 50g of water at 30C. Determine whether this solution is saturated. If yes, explain why.

3. A solution contains 5.2 g of potassium nitrate, KNO3, dissolved in 10 g of water at 40C. What amount of KNO3 would be required to saturate this solution?

4. A solution contains 33 g of KNO3/30g of water at 72C. How much must this solution be cooled to saturate the solution?


Solubility Curves Worksheet

Use the Solubility Curve on the previous page to help solve the following problems. Show your calculations where necessary.

1. Calculate the solubility of each of the following in g of solute/100g of water.a. 0.250 kg dissolves in 1.2L of water.

b. 24.0 g dissolves in 280.0 g of water.

2. Determine the solubility of the following in g of solute/100mL of water.a. 261 g of a solid dissolves in 1510mL of water.

b. 0.160 kg of a solid dissolves in 225 g of water.

3. Determine the temperature of the following substances using the given solubilitiesa. KNO3 120g/100g H2O

b. NaNO3 1200g/L H2O



4. What is the solubility, in g/100g H2O, of the following solutions at the specified temperature?a. NaNO3 at 40°C

b. Ce2(SO4)3 at 25°C

c. NH3 at 30°C

d. NH4Cl at 5°C

5. How much more NH4Cl can you dissolve in a litre of water at 60°C than at 20°C?

6. If you prepared a saturated solution of NaNO3 at 30°C and then cooled it to 8°C, what would happen? Be specific.

7. At which temperature do NaNO3 and KNO3 have the same solubility? What about NaCl and NH3?



8. How much water is needed to dissolve 65.0g of NaNO3 at 35°C?

9. What temperature is necessary to dissolve twice as much KNO3 as can be dissolved at 30°C?

10. If the solubility of a solid in water 118g/L of H2O, how much water would you need to dissolve a piece of the same solid with a mass of 45.0g?


Colligative Properties

The presence of solute particles makes the physical properties of a solution (i.e. boiling point, freezing point) ______________________________ from those of the pure solvent. These properties are known as colligative properties. It is the __________________________________ that influences the variation the most, not the chemical identity.

Boiling Point ElevationThe lowering of a solution’s vapour pressure elevates the boiling point above that of the pure solvent. Remember that the boiling point of a liquid is the temperature at which its vapour pressure equals

the atmospheric pressure. A solution ______________________ when it reaches the solvent’s boiling point because its

vapour pressure is still __________________________than the atmospheric pressure. A higher temperature is needed to raise its vapour pressure to equal the atmospheric pressure. The boiling point of an NaCl solution is ____________________________. Note that in general, only the ________________________________ vaporize; the solute

particles are ___________________________ in the solution.

Freezing Point Depression Molecules on the surface of the ice melt into water and molecules of water freeze on the surface

of the ice. When the ____________________________ is equal to the ___________________________, the

amount of ice and water is ___________________________. The melting point of water is 0˚C. This amount of water and ice will remain the same unless the conditions are changed in a way

that favours one of the processes. Adding salt (solute) to the ice and water makes it a solution. The salt __________________ some

of the water molecules surrounding the solid ice thereby blocking them from contacting the solid. The salt does not pack easily into the array of molecules in the solid. If the water cannot contact the ice they are unable to solidify. As a result fewer water molecules solidify, or the freezing rate _____________________________. In other words, the solution freezes at a lower temperature than the pure solvent. The freezing point of an NaCl solution is roughly -3.4˚C.

All aqueous solutions display this freezing point depression. The more salt present, the greater the freezing point depression. Any foreign substance (sugar, alcohol, etc.) will cause freezing point depression.

The practical applications of freezing point depression are numerous:1. Salt is used on roads and walkways to melt ice because it is inexpensive and readily available.2. Ethylene glycol is the main component of antifreeze and airplane de-icers.3. To survive in the Arctic, many fish and insects produce large amounts of glycerol, which lowers the

freezing point of their blood.


Concentration

It is important for chemists to know how much solute is in a certain volume of solution. When chemists speak about concentration they are referring to _____________________________________________________________. There are several different representations (units) for concentration.

Concentration : _____________________________________________________________________ _____________________________________________________________________

Dilute solution : _____________________________________________________________________

Concentrated solution : _______________________________________________________________

Representation of Concentration

1. Grams of solute per litre of solvent (g/L)

2. Percent weight of solute per volume of solution (%w/v)

3. Percent volume of solute per volume of solution (%v/v)

4. Parts per million (ppm)Represents the number of parts of one substance (solute) present in 1 000 000 (1 million) parts of another. It is used to represent small concentrations.

5. Parts per billion (ppb)Represents the number of one substance (solute) present in 1 000 000 000 (1 billion) parts of another. It is used to represent very small concentrations. The Public Health Service just reduced the permissible lead level in drinking water from 50 ppb to only 15 ppb. Even miniscule levels of lead may cause nerve damage and learning disabilities.

6. _________________________________ (most commonly used)Represents the number of moles of solute dissolved in 1L of solution.The units are mol/L or mol/dm3 because 1 L = 1 dm3

The volume is the volume of the total solution, not just the solvent.A one-molar (1M) solution of NaCl contains 1 mole of NaCl per 1L of solution.

Note: 1 L = 1 dm3 = 1000 mL = 1000 cm3


Molarity/Concentration

The units for concentration are _____________________. This is also known as molar concentration, or molarity, M. Concentration tells you how many moles of solute are in one litre of a solution or mathematically,

concentration (M) = moles of solute (mol)litres of solution (L)

C= nV

Example: What is the concentration of 52.0 g of NaCl dissolved in 200.0 mL of water?

*always find number of moles first*

Example: What is the concentration of a solution in which 0.45 g of sodium nitrate is dissolved in 265 mL of solution?

Example: What will the volume of a 0.50 M solution be if it contains 25 grams of calcium hydroxide?


Molarity Calculations Worksheet

Solve each of the following problems. Show your work and include units.

1. Calculate the molarity of the following solutions.a. 12 g of lithium hydroxide (LiOH) in 1.0 L of solution

b. 198 g of barium bromide (BaBr2) in 2.0 L of solution

c. 54 g of calcium sulfide (CaS) in 3.0 L of solution

2. Calculate the volume of each solution, in litres.a. 1.0 M solution containing 85 g of silver nitrate (AgNO3)

b. 0.5 M solution containing 250 g of manganese (II) chloride (MnCl2)

c. 0.40 M solution containing 290 g of aluminum nitrate (Al(NO3)3)

1a.) 0.50 M b.) 0.33 M c.) 0.25 M 2a.) 0.50 L b.) 3.97 L c.) 3.40 L


Molarity and Stoichiometry Worksheet

1. Calcium hydroxide and sulphuric acid react to produce calcium sulphate and water according to the following balanced equation:

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

a. How many litres of 0.50 M calcium hydroxide do you need in order to have 5.5 moles of calcium hydroxide?

b. Find the number of moles of sulphuric acid needed to react with 5.5 moles of calcium hydroxide.

c. If the sulphuric acid has a concentration of 0.82 M, how many litres of it are needed to react with 5.5 moles of calcium hydroxide?

2. Calcium carbonate reacts with hydrochloric acid according to the following balanced equation:

CaCO3(aq) + 2HCl(aq) CO2(g) + CaCl2(aq) + H2O(l)

a. What mass of calcium carbonate is needed to make 1.20 litres of a 1.70 M calcium carbonate solution?

b. What volume of 3.00 M hydrochloric acid is needed to completely react with the amount of calcium carbonate in 2a above?

c. Based on 2a and 2b above, how many moles of water would be produced?

1a.) 11 L b.) 5.5 mol c). 6.71 L 2a.) 204.18 g b.) 1.36 L c.) 2.04 mol


Molarity and Stoichiometry Worksheet

3. Ammonium chloride and calcium hydroxide react according to the following balanced equation: 2 NH4Cl(aq) + Ca(OH)2(aq) CaCl2(aq) + 2 NH3(g) + 2 H2O(l)

a. What mass of ammonium chloride is needed to make 1.0 L of a 2.0 M ammonium chloride solution?

b. What mass of calcium hydroxide is needed to make 2.00 L of a 2.00 M calcium hydroxide solution?

c. How many grams of calcium chloride will be made when 1.00 L of a 1.00 M calcium hydroxide solution react with excess ammonium chloride?

4. Zinc and hydrochloric acid react according to the following balanced equation:

Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2(g)

a. What volume of 0.1 M hydrochloric acid will react with 26 g of Zn?

b. What mass of zinc will react with 2.0 L of 0.25 M hydrochloric acid?

c. How many litres of hydrogen will you make (at STP) if you react 2.74 L of 0.450 M hydrochloric acid with excess zinc?

3a.) 106.98 g b.) 296.4 g c.) 110.98 g 4a.) 7.95 L b.) 16.35 g c.) 13.81 L


Preparing a Solution In making a solution with a solid solute and liquid solvent, chemists follow a few basic steps:

1. ________________________________________________________________________________

2. ________________________________________________________________________________

3. ________________________________________________________________________________

________________________________________________________________________________

4. ________________________________________________________________________________

________________________________________________________________________________

5. ________________________________________________________________________________

Example: Describe the steps needed to make 500.0 mL of a 0.100 M solution of sodium hydroxide.


Preparing a Solution Example:What mass of sodium chloride is needed to make 0.250 L of a 0.100 M solution?

Example:What mass of copper (II) sulphate is needed to make 600 mL of a 0.500 M solution?


Preparing a Solution Worksheet

1. What mass of AgNO3 would be needed to make 250.0 mL of a 1.50 M solution?

2. What mass of CaCO3 would be needed to make 20.0 mL of a 0.400 M solution?

3. How much solution would be needed to dissolve 50.0 g of K2SO4 to make a 0.500 M solution?

4. What volume of solution would be required to dissolve 18.04 g of aluminum sulphide to make a 0.160 M solution?

5. What is the chloride ion concentration in a 0.25 M solution of iron (III) chloride?

6. What mass of cobalt (III) nitrate is needed to make 1.25 L of a solution with a nitrate ion concentration of 0.150 M?

1.) 63.71 g 2.) 0.801 g 3.) 0.57 L 4.) 0.75 L 5.) 0.75 M 6.) 15.31 g


Dilution of a Solution

A concentrated solution has a lot more solute as compared to solvent. A dilute solution has little solute as compared to solvent. Concentrated and dilute are relative terms. This can be illustrated by the three solutions below:

Solution A Solution B Solution C500 g of sugar in 100 mL of water 100 g of sugar in 100 mL of water 1 g of sugar in 100 mL of water

Solution A is __________________ since it has a large amount of solute compared to solvent. Solution C is ______________ because it has a small amount of solute compared to solvent. Solution B is dilute compared to solution A since it has less solute, but is concentrated

compared to solution C because it has more solute than solution C.

Dilutiono _________________________________________________________________________o _________________________________________________________________________

number of moles of solute before dilution = number of moles of solute after dilution

o _________________________________________________________________________o The concentration changes according to the following equation:

Example: Calculate the final concentration when 75.0 mL of water is added to 25.0 mL of 8.00 M HCl.

Example: What volume of water must be added to 150.0 mL of a 5.0 M solution of NaOH to make a 2.00 M solution?

Example: Calculate the volume of stock 18.0 M H2SO4 that would be required to make 300.0 mL of 3.00 M solution.


C1V1 = C2V2

Dilutions Worksheet

1. If I add 25 mL of water to 125 mL of a 0.150 M NaOH solution, what will the molarity of the diluted solution be?

2. If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be?

3. How much 0.05 M HCl solution can be made by diluting 250 mL of 10 M HCl?

4. I have 345 mL of a 1.5 M NaCl solution. If I boil the water until the volume of the solution is 250 mL, what will the molarity of the solution be?

5. How much water would I need to add to 0.50 L of a 2.4 M KCl solution to make a 1.0 M solution?

1.) 0.13 M 2.) 0.10M 3.) 50 L 4.) 2.07 M 5.) 0.70 L


Documents

UNIT 4 – Solutions · Web viewChemistry 30S Solutions Unit Four Describe and give examples of various types of solutions. Describe the structure of water in terms of electronegativity