unit 10: solutions-lecture regents chemistry ’14 mr ... · unit 10: solutions-lecture regents...

Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch

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Unit 10:

Solutions

Student Name: _______________________________________

Class Period: ________



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Unit 10 Vocabulary:

1. Aqueous: A solution in which the solvent is water.

2. Colligative Property: A property of a solution that is dependent on

concentration. Examples include boiling point, freezing point, and

vapor pressure.

3. Molality: The concentration of a solution measured in moles of solute

per kilogram of solvent.

4. Molarity: The concentration of a solution measured in moles of solute

per liter of solution.

5. Parts per Million: The concentration of a solution measured in mass

of solute per mass of solution multiplied by one million.

6. Percent by Mass: The concentration of a solution measured in mass of

solute per mass of solution multiplied by one hundred.

7. Percent by Volume: The concentration of a solution measured in

volume of solute per volume of solution multiplied by one hundred.

8. Saturated: Any solution that has the maximum concentration of a

dissolved solute possible in a given quantity of solvent at a given

temperature. A saturated solution is a solution at equilibrium.

9. Solubility: The maximum quantity of a solute that may be dissolved

in a given quantity of solvent at a given temperature to make a

saturated solution.

10. Solute: Any substance that is broken apart by a solvent and kept

separate by the solvent particles.

11. Solution: A homogenous mixture formed when a solute dissolves

into a solvent.

12. Solvent: Any substance that attaches to solute particles, breaks the

solute apart, and then keeps the separated particles apart in solution.



13. Supersaturated: A solution that has an excess of solute beyond the

solubility point for a given temperature. Excess solute will either

precipitate out, or remain in an unstable dissolved state until the

supersaturated solution is disturbed. This causes the excess solute to

precipitate from solution, leaving the solution saturated.

14. Unsaturated: A solution in which there are solvent particles that

have no attached solute particles, and therefore has the capacity to

take more solute into solution.



Unit 10 Homework Assignments:

Assignment: Date: Due:



Notes page:



Mixtures:

A mixture is made by physically combining two (or more)

substances without any type of chemical reaction occurring.

Mixing ionic compounds with water forms aqueous solutions

composed of dissolved ions. The polar water molecules attach to the

ions, separating them from other ions. The polar water molecules

keep the ions separate, holding the ions apart. This property is called

molecule-ion attraction.

Topic: Mixtures

Objective: What is the result of combining two dissimilar substances?



A solid crystal of NaCl is placed into a beaker of water. Immediately

the polar water molecules attract the charged ions. The partially

positive ends of the water molecules (hydrogen atoms) attract the

negatively charged chloride ions. The partially negative ends of the

water molecules (oxygen atoms) attract the positively charged sodium

ions. When enough water molecules attach to an ion, the natural

kinetic motion of the liquid water molecules will “tear” the ions

from each other. This is the molecule-ion attraction.

Topic: Molecule-Ion Attraction

Objective: How are ions situated within a solution?



The hydrated (aqueous) ions are kept separated by the motion of the

water molecules that are attracted to them. In this example, four

water molecules around each ion are enough to keep the ions

separated. There are still “free” water molecules in the solution

which could “tear” apart, attract, and keep separate additional solute

ions. This is an example of an unsaturated solution. If more NaCl(s)

were added until all water molecules were attached to ions, the

solution would then become saturated. At the saturation point, any

additional NaCl(s) added would simply sink to bottom of the beaker

without going into solution, or drive other Na+1

or Cl-1

ions out of

solution as a precipitate. The number of water molecules required to

keep ions apart depends on water temperature. The higher the

temperature, the greater the average kinetic energy, and the faster the

water molecules move, so fewer water molecules would be required

to keep the ions apart.



Solubility:

The quantity of a solute that may be added to a given quantity of

solvent at a given temperature and pressure is known as solubility.

Saturated: A saturated solution holds as many dissolved particles as

it has the capacity to hold. Capacity is the maximum “spaces”

available for anything. The chemistry room has 23 desks; this gives

Room 218 the capacity for 23 students unless more desks are brought

in, or if desks are taken out. In a saturated aqueous solution, all the

solvent water molecules (desks) have an ion (student butt) attached to

them, so no more ions (students) could come into the solution (Room

218) and find a place to stay (empty desk). Any additional ions

(students) would need to pass through the solution (room).

Unsaturated: An unsaturated solution holds fewer dissolved

particles as it has the capacity to hold. Room 218 has 23 desks, and

the largest chemistry class (for 2014-2015) has 21 students, so there is

capacity for two more students. In an unsaturated aqueous solution,

there are solvent water molecules (desks) without an ion (student

butt) attached to them, so more ions (students) could come into the

solution (Room 218) and find a place to stay (empty desk).

Topic: Solubility

Objective: How may we know the amount that enters into solution?



Supersaturated: A supersaturated solution is very rare solution in

where the solution holds more dissolved solute than is theoretically

possible. A supersaturated solution is an unstable solution that will

cause the excess solute to leave (precipitate) if disturbed. The

solution (Room 218) has 23 desks (water molecules) that each can

attach to one ion (student butt). If there are more than 23 students

(ions) sitting, whereas some desks have more than one student (ion)

sitting in them, this is over capacity, or supersaturated. The solution

can easily be upset if a disturbance (Murdoch) comes in contact with

the system, and forces the extra ions (students) out of the solution

(Room 218).

Factors affecting solubility:

Three factors affect solubility:

1. Temperature

2. Pressure

3. Nature of solute and solvent

Watch Crash Course Chemistry Solutions - YouTube - 8:20

https://www.youtube.com/watch?v=9h2f1Bjr0p4&index=27&list=PL8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr



Temperature:

1. For solid and liquid solutes, solubility in water increases as water

temperature increases.

2. For gaseous solutes, solubility in water decreases as water

temperature increases.

Effect of temperature on solubility of a solid solute in water:

An example of the effect of temperature on aqueous solubility of

solids is the sugar water needed to form rock candy. A saturated

solution is formed at high temperature. As the saturated sugar

solution cools, the sugar solute becomes less soluble, causing some

water molecules to “jump” off a sugar molecule and assist other water

molecules in holding other sugar molecules apart. This allows the

now “free” sugar molecules to come out of solution, and as more

water molecules are pulled off sugar molecules, more and more sugar

will precipitate from the solution as sugar rock candy.

Topic: Temperature and Solubility

Objective: How does temperature affect the solubility of a solution?



Effect of temperature on solubility of a gaseous solute in water:

An example of the effect of temperature on aqueous solubility of

gases is as easy to find as your favorite carbonated beverage. Take

two identical bottles of the same carbonated beverage, place one in

the refrigerator overnight, and leave the other on the kitchen counter.

When you open both bottles side by side you will see that the colder

beverage will release much less carbonation (bubbles) than will the

warmer beverage. The colder solution allowed more gaseous CO2 to

remain dissolved than did the warmer solution. If you keep observing

both bottles, as the colder beverage warms, it will continue to lose

dissolved CO2 (bubbles) during the warming process, while the

warmer bottle will form fewer bubbles during the same elapsed time.



Pressure:

For solids and liquids, pressure has either no effect or a negligible

effect on solubility.

For gaseous solutes, solubility increases as pressure increases.

i. In commercial soda machines, as water travels through a tube,

flavored syrup is added to the water along with carbon dioxide gas.

Gases don’t have an affinity to form aqueous solutions and require

pressure to force the CO2 gas into the water and syrup solution.

When the aqueous syrup and CO2 mixture is bottled, the gas is

“trapped” within the volume of the container, and equilibrium

between dissolved CO2 molecules and free CO2 molecules quickly

forms. When the container is opened, the pressure decreases,

allowing the CO2 to escape as bubbles. CO2 gas is soluble at high

pressures (sealed container) but nearly insoluble at low pressure

(open container).

Topic: Pressure and Solubility

Objective: How does pressure affect solubility of solutes?



Nature of Solute and Solvent (Like dissolves Like):

Polar solutes dissolve in polar solvents.

o Water is a polar molecule and therefore has partially charged ends

that can attract other polar or ionic solutes. This is why ionic

solutes such as NaCl and polar molecular solutes such as glucose

(sugar) may be dissolved in water.

o Water’s polar structure has little attraction for nonpolar molecular

solutes like oil (remember Hank squishing butter?) Oils will not

mix (are immiscible) with water, and being less dense, will float on

top of the water. This is why water is NOT used to extinguish oil

fires; the water will go under the fire, and float the fire to a new

location. Nonpolar gases such as CO2 and O2 are even harder to

force into aqueous solution because of this.

Nonpolar solutes dissolve in nonpolar solvents.

o Oils will not dissolve in polar water solvent, but oils will easily

dissolve in the nonpolar organic solvent benzene (C6H6). Whereas

water is a great polar solvent, benzene is a good nonpolar solvent.

Acetone is a nonpolar solvent used to remove nonpolar mixtures

such as nail polish.

Topic: Polar Nature of Solvent

Objective: Does the polarity of a solute or solute apply in solutions?



Solubility Regents Practice Problems (ungraded):

1. At room temperature, the solubility of which solute listed below would the most

affected by a change in pressure?

a) Sugar(s)

b) Methanol(l)

c) Sodium nitrate(aq)

d) Carbon dioxide(g)

2. As the pressure on a gas confined above a liquid increases, the solubility of the

gas in the liquid

a) Increases

b) Decreases

c) Is unchanged

3. Sublimated carbon dioxide is most soluble in water under the conditions of

a) Low pressure and low temperature

b) High pressure and low temperature

c) Low pressure and high temperature

d) High pressure and high temperature

4. At which temperature given below could water contain the most dissolved gas

at a pressure of 101.3 kPa?

a) 183˚C

b) 283 K

c) 383 K

d) 483˚C



Reference Table G: Solubility Curves at Standard Pressure:

1. For Table G, the SLOPE of the curve tells you about the solute!

a. IONIC salts solutes have UPWARDS slopes;

b. GAS solutes (SO2 and NH3) have DOWNWARDS slopes.

2. Interpreting Reference Table G:

a. The Table G x-axis is the temperature in ˚C. If you are given a

temperature in K, you’ll need to convert (K - 273 = ˚C). Note that

the numbers are significant to TWO places, as all numbers have a

decimal (.) at the end. Note also that the interval is to the tens

place (10.˚C), so precision may be interpolated to the ones place

and will still be significant to TWO digits.

b. The Table G y-axis is the solubility in grams of solute per 100.

grams of water. Again, the numbers are significant to TWO places,

as all numbers have a decimal (.) at the end. Note also that the

interval is to the tens place (10. grams), so precision may be

interpolated to the ones place and will still be significant to TWO

digits.

c. Also note the segment of the curve for potassium iodide WAY up

in the extreme upper left corner. Students have missed the KI

curve; be aware of its presence BEFORE you need it!

Topic: Solubility Curves

Objective: How may we use solubility curves to determine solubility?



3. Table G is only useable at standard pressure (1 atm or 101.3 kPa); if

you have other than standard pressure, you’ll need to infer.

Solubility of a solute in 100. grams of water at various temperatures:

1. Each line on Table G represents a saturated aqueous solution of the

given solute for a given temperature. The farther up the y-axis the

saturation line is at a given temperature, the more soluble the solute

is. To find the saturation point at a given temperature, simply find the

given temperature and then go straight vertically up until you

intersect the solubility line you need, and then traverse level left and

read the solubility on the y-axis. The table is set for grams of

solute/100. grams of H2O; if you have other than 100. g of water you

will need to find the correct proportion.



Determining Molecular Polarity:

Degree of saturation according to Table G:

1. Unsaturated:

2. Saturated:

Topic: Degree of Saturation

Objective: How do we use solubility curves to determine saturation?



3. Supersaturated:



Solubility for other than 100. g of water solvent:

Reference Table G may only be directly used in the mass of the

solute is dissolved in 100. grams of water as a solvent. With simple

math, by finding the proportionality of the given amount of solvent to

100. g of water as listed on Table G, we can easily calculate the

solubility at the given mass of water.

If given a question with OTHER than 100. grams of water:

1. For the stated temperature in the question, find the saturation

solubility in grams of solute per 100. grams of water.

2. Place the question’s stated amount of water as a numerator and

100. g of water as the denominator. This will give you a

multiplier.

i. If you had MORE than 100. grams of water in your question,

the multiplier factor will be greater than 1.

ii. If you had LESS than 100. grams of water in your question, the

multiplier factor will be less than 1.

3. Multiply the saturation solubility from Table G by the multiplier

factor from Step #2, and you will have your saturation solubility

for other than 100. grams of water.

Topic: Solubility in other than 100. g

Objective: Will Table G determine solubility not in 100. g of water?



I. How many grams of KClO3 are required to make a saturated

aqueous solution in 100. grams of water at 30.°C?

a. Find 30.°C on the Table G x-axis, then move straight up to the

KClO3 curve. Move straight over to the y-axis, and read the scale

(12 g). This means that 12 grams of KClO3 are soluble in 100.

grams of water at 30.°C.



II. How many grams of KClO3 are required to make a saturated


a. We are now asked for the solubility of KClO3 in 50. grams of

water. Place 50. g of water in the numerator, and 100. g of water

in the denominator, and solve.

50. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

100. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 0.50 multiplier

b. We used Table G to determine that 12 g of KClO3 are soluble in

100. g of water. We can use the multiplier factor with the 12 g of

KClO3 and determine the mass of KClO3 soluble in 50. g of

water.

c. 12 g of KClO3 x 0.50 = 6.0 g of KClO3 soluble in 50. g of water at

30.°C

III. How many grams of KClO3 are required to make a saturated


a. We are now asked for the solubility of KClO3 in 200. grams of

water. Place 200. g of water in the numerator, and 100. g of water

in the denominator, and solve.

200. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟


b. We used Table G to determine that 12 g of KClO3 are soluble in

100. g of water. We can use the multiplier factor with the 12 g of

KClO3 and determine the mass of KClO3 soluble in 200. g of water.

c. 12 g of KClO3 x 2.00 = 24 g of KClO3 soluble in 200. g of water at

30.°C



IV. Let’s say you have to make a saturated NaCl solution using tap

water (10°C) but you only have an empty (355 mL) soda can. Of

course you know that water has a density of very close to 1.0

gram/mL near 10°C, so you can use the 355 mL volume of your

soda can to measure out 355 grams of water. How much NaCl will

be needed to form the saturated solution with the 10°C water?

a. First, start at 10°C on Table G and go up until you find the

solubility curve for NaCl (blue dashed line), and then move left

until you reach the y-axis scale (green dotted line). I’ll call it 37.

However, that 37 g of NaCl is for 100 grams of water! Since you

have 355 grams of water, 355 g needs to go in the numerator, with

the Table G standard of 100. grams of water as the denominator.

355 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟


b. We can now take the interpolated 37 grams of NaCl per 100.

grams of water and multiply that by 3.55 to give us the saturation

point of NaCl in 355 grams of water.

37g



Table G example questions:

1. What is the solubility of HCl in 100. g of water at 70˚C?

a. Find 70˚C on the x-axis, move straight up to the HCl curve, and

then transit level left to the y-axis.

b. 52 g of HCl are soluble in 100. g of water at 70˚C

2. What type of solution saturation do you have if 20. g of KClO3 are

dissolved in 100. g of water at 40.˚C?

a. Find 40.˚C on the x-axis, move straight up to the KClO3 curve, and

then transit level left to the y-axis.

b. 16 g of KClO3 are soluble in 100. g of water at 40.˚C. As you

were asked what 20. g of KClO3 in solution would be, since 20 is

greater than 16, the solution would be supersaturated.

3. For the question above, how can we make the existing solution

saturated?

a. As calculated using Table G, at 40˚C only 16 g of KClO3 could be

dissolved to make the solution of aqueous KClO3 saturated. With

the stated 20. g of KClO3, it was supersaturated. We can make the

solution saturated if we increase the temperature to around 48˚C,

the point that the curve crosses the 20. g solute/100. g H2O level.

b. What happens in the example if we increase the temperature of the

aqueous solution to 75˚C without adding more solute? When we

find 75˚C on the x-axis and go straight up to the curve, we see that

75˚C crosses the curve at 40. g solute/100. g H2O level. That

would mean that with a capacity of 40. g, and a dissolved solute

amount of 20. g, the aqueous solution of 20. g per 100. g of water

would only be about 50% saturated at 75˚C, or about 50%

unsaturated.



Solubility Regents Practice Problems (ungraded):

1. Solubility data for four different salts aqueous solution at 60˚C are given in the

data table below.

Which given salt is the most soluble at 60C?

a) Salt A b) Salt B c) Salt C d) Salt D

2. According to Reference Table G, which of these substances is most soluble at

60˚C?

a) Sodium chloride

b) Potassium chloride

c) Potassium chlorate

d) Ammonium chloride

3. According to Reference Table G, how many grams of potassium nitrate would

be needed to saturate 200 grams of water at 70˚C?

a) 43 g

b) 86 g

c) 134 g

d) 268 g

4. According to Reference Table G, how does decreasing the average kinetic

energy affect the solubility of NH3 and KCl?

a) The solubility of NH3 increases, and the solubility of KCl increases.

b) The solubility of NH3 increases, and the solubility of KCl decreases.

c) The solubility of NH3 decreases, and the solubility of KCl increases.

d) The solubility of NH3 decreases, and the solubility of KCl decreases.



Student name: _________________________ Class Period: _______

Please carefully remove this page from your packet to hand in.

Solutions Homework

Complete the following the chart: (1 pt. ea.)

For the

solutes

below, in

aqueous

solution:

If temperature of the

solvent is increased,

the solubility of this

solute will:

If the temperature of

the solvent is

decreased, the

solubility of this solute

will:

If the surface area of

this solute is

increased, the

solubility of this

solute will:

NH3(g)

KCl(s)

Complete the following chart: (1 pt. ea.)

Sketch the structures of

the compounds.

Is the solute polar,

nonpolar, or ionic?

Will this solute dissolve in

water or benzene?

CH4

KBr

H2S

Explain why, in terms of molecular polarity, why oils will not easily dissolve in

water. (1 pt.)

Cont’d next page



Use Reference Table G to answer the following: (1 pt. ea.)

________ 1. As the temperature of water decreases, the solubility of gases will?

________ 2. As temperature of water decreases, the solubility of solids will?

________ 3. Which salt on Table G is the least soluble at 90.˚C?

________ 4. Which gas on Table G is the most soluble at 60.˚C?

________

5. Which salt on Table G shows the least change in solubility between

30.˚C and 70.˚C?

For #6 through #8, determine whether the following solutions are Unsaturated,

Saturated, or SuperSaturated in 100. grams of water.

________ 6. 108 g of KNO3 at 60.˚C

________ 7. 20. g of NH4Cl at 30.˚C

________ 8. 45 g of KCl at 60.˚C

________ 9. What is the solubility of NaNO3 in 100. grams of water at 40.˚C?

________ 10. How much KClO3 may be dissolved in 1000. g of water at 30.˚C?

________ 11. What is the solubility of NaCl in 50. g of water at 70.˚C?

________

12. How many grams of NH4Cl may be added to a solution containing

30. g of NH4Cl to make a saturated solution in 90.˚C H2O?

________

13. A solution has 80. g of KNO3 dissolved in 80.˚C H2O. At what

temperature would the solution become saturated?

________

14. A saturated solution of NaNO3 cools from 343 K to 293 K. If the

solution stays saturated, how many grams of NaNO3 precipitate?

________

15. A saturated solution made in 50. g of water at 303 K of NaNO3 is

left out and fully evaporates. What is the mass of NaNO3 left?

________ 16. At what temperature will KCl and HCl have the same solubility?



37 g of NaCl x 3.55 = 131.55, or about 130 grams of NaCl are soluble in

355 grams of water at 10°C

Concentration:

The concentration of a solution is a measure of the amount of

substance per unit of volume.

a. When discussing solutions, there are several ways to express

concentration.

1. Grams of solute/100 g of solvent (water):

This is how Reference Table G compares solubility.

2. Molarity (M):

Molarity is the number of moles of solute per liter of solvent.

Molarity is the most common unit of concentration used in the

laboratory.

Topic: Concentration

Objective: How do we express the amount of solute per unit volume?



Determining Molarity Experimentally:

We can experimentally determine the molarity of a solution by

1. Measuring the volume of the liquid solution you have

2. Determine the number of moles of solute in the solution

3. Evaporate all solvent leaving only solute

4. Mass the recovered solute

5. Divide the measured mass of recovered solute by the gram formula

mass of the solute

Topic: Determining Molarity

Objective: How do we work with a single element and compound?



Molarity examples:

1. What is the molarity of a solution containing 2.0 moles of KNO3

dissolved in 4.0 L of water?

a. M = moles of solute/L of solution 2.0 𝑚𝑜𝑙𝑒𝑠 𝐾𝑁𝑂3

4.0 𝐿 𝑤𝑎𝑡𝑒𝑟 = 0.5 M KNO3

2. What is the molarity of a solution containing 2.0 moles of HCl

dissolved in 250. mL of water?

a. As molarity is in units of liters (L) of solution, first convert mL to L.

b. 250 mL x 1 𝐿

1000 𝑚𝐿 = 0.250 L

c. M = moles of solute/L of solution 2.0 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝑙

0.250 𝐿 𝑤𝑎𝑡𝑒𝑟 = 8.0 M HCl

3. What is the molarity of a solution containing 20. grams of NaOH

dissolved in 2.0 L of water?

a. As molarity is in units of moles of solute, first convert grams of

NaOH to moles of NaOH after finding the gram formula mass for

NaOH.

b. NaOH = (23.0 + 16.0 + 1.0) = 40.0 g / mole NaOH

c. Then convert 20. grams of NaOH to moles of NaOH.

d. 20. g NaOH x 1 𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐻

40.0 𝑔 𝑁𝑎𝑂𝐻 = 0.5 mole NaOH

e. M = moles of solute/L of solution 0.5 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻

2.0 𝐿 𝑤𝑎𝑡𝑒𝑟 = 0.25 M NaOH

KNO3



4. What is the molarity of a solution containing 60. grams of NaOH

dissolved in 400. mL of water?

a. As molarity is in units of moles of solute, first convert grams of

NaOH to moles of NaOH.

b. 60. g NaOH x 1 𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐻

40.0 𝑔 𝑁𝑎𝑂𝐻 = 1.5 mole NaOH

c. As molarity is in units of liters (L) of solution, first convert mL to

L.

d. 400 mL x 1 𝐿

1000 𝑚𝐿 = 0.400 L

e. M = moles of solute/L of solution 1.5 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻

0.400 𝐿 𝑤𝑎𝑡𝑒𝑟 = 3.75 or 3.8 M

NaOH



The basic equation for molarity of molarity = moles of solute/liter of

solution (Reference Table T) may be rearranged as needed.

i. To make a specific volume of a solution with a desired molarity:

1. If you are given a volume and a molarity, you need to solve for

moles of solute.

2. Moles of solute = M (molarity) x L (volume of solution)

ii. Steps to make your solution:

1. Determine the desired concentration (what is the molarity of the

solution you are trying to make? This is usually told to you in the

lab handout.)

2. Determine how much solution you want to make (how many L or

partial L of solution you need? This, again, is usually told to you

in the lab handout.)

3. Multiply the first two step’s answers. This will give you your

amount of solute in moles.

4. 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 x L of solution = moles of solute

5. Calculate the gram formula mass for your required solute, to the

nearest tenth of a gram/mole.

Topic: Calculating Molarity in Lab

Objective: How do we make a solution with a desired molarity?



6. Multiply the number of moles of solute needed from step 3 by the

gram formula mass of your solute.

7. # of moles calculated x 𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 𝑖𝑛 𝑔

1 𝑚𝑜𝑙𝑒 = grams of solute needed

8. Measure your calculated grams of solute.

9. Add your solute to a half-full volumetric flask that matches your

needed volume.

10. When the solute is dissolved, fill the flask to the needed volume.

Wow. Seems like a LOT of work, doesn’t it? Why bother? Who would

want to do all that?

Well,

I (or any other chemistry teacher, lab technician, nurse, doctor,

veterinarian, etc.) have to do ALL of those steps frequently. Creating a

solution to a specific molarity is one of the most commonly done tasks

in chemistry. Once you understand the basics, you will be able to do it

for life.



Examples for calculating molarity:

1. How many grams of NaOH are needed to make 4.0 L of 0.50 M

aqueous NaOH?

a. Determine the number of moles of NaOH needed:

Moles = M x L (0.50 moles/L) x (4.0 L) = 2.0 moles of NaOH

b. Convert moles of solute to grams of solute:

Grams = moles of solute x gram formula mass of solute = (2.0

moles NaOH) x (40.0 g NaOH/mole NaOH) = 80. g of NaOH

c. You can now measure and then add 80. grams of NaOH to a

partially filled 4.0 L volumetric flask and add distilled water to the

4.0 L level.

2. How many grams of KCl are needed to make 500. mL of 0.100 M

aqueous KCl?

a. Molarity is in units of liters (L) of solution, first convert mL to L.

b. 500. mL x 1 𝐿

1000 𝑚𝐿 = 0.500 L

c. Determine the number of moles of KCl needed:

Moles = M x L (0.100 moles/L) x (0.500 L) = 0.0500 moles of

KCl

d. Calculate the gram-formula mass of KCl:

KCl = (39.1 + 35.5) = 74.6 g KCl/mole

e. Convert moles of solute to grams of solute:

Grams = moles of solute x gram formula mass of solute = (0.0500

moles KCl) x (74.6 g KCl/mole KCl) = 3.73 g of KCl

f. You can now measure and then add 3.73 grams of KCl to a partially

filled 500. mL volumetric flask and add distilled water to the 500.

mL level.



Parts-per-million (ppm):

Parts-per-million is a mass-to-mass concentration for determining

impurities or pollutants.

Parts-per-million is used to determine trace amounts of ions in

drinking water. PPM may also be used to measure atmospheric or

industrial air pollutants. For some very toxic materials concentrations

may be expressed in parts-per-billion.

i. What is the ppm concentration of lead ions when 0.0000450 g of

lead is dissolved in 100. grams of water?

Ppm = ( 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ) x 1,000,000 = (

0.0000450 𝑔 𝑜𝑓 𝑃𝑏

100. 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ) x 1,000,000

= 0.45 ppm Pb

Topic: Parts-per-Million

Objective: How would we calculate the concentration in mass?



Ppm examples:

1. A well drilled on property near a landfill was tested before a housing

development was constructed. A 1.000 kg sample of the well water

was found to have 0.00330 g of cadmium, a toxic heavy metal found

in rechargeable batteries.

a. What is the ppm concentration of cadmium in the well water?

Ppm is measured in grams of solution, convert kg to grams:

kg x 1000. 𝑔

1.000 𝑘𝑔 = 1000. grams

Ppm = 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 x 1,000,000

0.00330 𝑔 𝑜𝑓 𝐶𝑑

1000. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 x 1,000,000 =

3.30 ppm of cadmium

2. What would be ppm concentration of a solution containing 5.00 g of

solute dissolved in 250. g of water?

For this example you are given the components of the solution, not

the total mass of the solution. We need to add both components

(solute and solvent) together.

5.00 g of solute + 250. g of water = 255 g of solution

Ppm = 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 1,000,000

5.00 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

255 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 1,000,000 =

19607.8 ppm 19,600 ppm



Percent by Mass:

Percent by Mass = (grams of solute/grams of solution) x 100

i. A 25.0 g sample of aqueous NaCl solution is evaporated and 2.0 g of

NaCl crystals are recovered. What is the percent by mass of the NaCl

in the original solution?

ii. Percent by Mass = 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100 =

2.0 𝑔 𝑜𝑓 𝑁𝑎𝐶𝑙

25.0 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100 =

8.0% NaCl by mass

Percent by Volume:

Percent by Volume = (mL of solute/mL of solution) x 100

i. Percent by volume is often used to describe the concentration of

alcohol. Whiskeys are supposed to be 50% ethanol (EtOH) or more;

50% EtOH(aq) is the point where it remains flammable. A customer

thinking it was too watered down might ask for proof it wasn’t

watered-down, and if it didn’t burn, it was not strong enough. This is

the basis for the term “proof” in comparing distilled spirits.

ii. A 50.0 mL sample of an aqueous ethanol solution is distilled to yield

33.2 mL of EtOH. What percent by volume of EtOH was the original

solution?

Percent by volume = 𝑚𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑚𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100

33.2 𝑚𝐿 𝑜𝑓 𝐸𝑡𝑂ℎ

50.0 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100

= 66.4% ethanol by volume

Topic: Percent by Mass & Volume

Objective: Can we calculate concentration as a function or percent?



Concentration Regents Practice Problems (ungraded):

1. What is the molarity of a solution containing 20. grams of NaOH in 500 mL of

water?

a) 1 M

b) 2 M

c) 0.5 M

d) 0.04 M

2. How many moles of solute are in 200 mL of a 1 M solution?

a) 1

b) 0.2

c) 0.8

d) 200

3. What is the concentration of a solution containing 10. moles of copper(II)

nitrate in 5000. mL of water?

a) 1.0 M

b) 2.0 M

c) 5.0 M

d) 0.50M

4. What is the total number of grams of NaI(s) needed to make 1.0 liter of a 0.010

M aqueous solution?

a) 15

b) 1.5

c) 0.15

d) 0.015

5. Using your reference tables, which compound listed would form the most

concentrated aqueous solution?

a) AgBr

b) AgCl

c) AgNO3

d) Ag2CO3

6. What is the concentration of a solution, in parts per million, if 0.02 gram of

Na3PO4 is dissolved in 1000. grams of water?

a) 2 ppm

b) 20 ppm

c) 0.2 ppm

d) 0.02 ppm



Notes page:





Concentrations Homework

You must show ALL work, including numerical setup, units, and correctly rounded

answers. (4 pt. ea.)

1. Calculate the molarities of the following solutions:

a) 2.0 moles of NaCl(s) in 4.0 liters of solution

b) 34 grams of CaSO4(s) in 2000. mL of solution

c) 28 grams of KOH(s) in 100. mL of solution

2. Calculate the mass of solute needed to make the following solutions:

a) 1.0 L of 2.0 M NaOH(aq)

b) 500. mL of 10.0 M HCl(aq)

c) 1.0 L of 5.0 M Ba(NO3)2(aq)

3. 2.5 grams of NaOH(s) were recovered after 50.0 mL of NaOH(aq) completely

evaporated. What was the molarity of the original solution?

4. Calculate the percent by mass of 35.8 g of Na2SO4(aq) in 136.3 grams of

solution.



5. Calculate the percent by volume of 4.55 mL of ethanol in 7.83 mL of solution.

Parts Per Million: (Show ALL work)

6. A sample is found to contain 0.015 g of Pb+2

in 10 grams of tap water.

Calculate the concentration of the lead in ppm for this sample.

7. The tap water analyzed above came from a site near a landfill. The EPA tested

groundwater samples near the landfill, and found the average groundwater

sample around the landfill contained 2.7 x 10-4

grams of Pb+2

per kilogram of

water. State your conclusion if the lead in the tap water came from the landfill,

and back up the answer with your calculations.

8. A 100.0 mL sample of 8.50 ppm aqueous solution completely evaporates.

What will be the mass of the recovered solute?

9. A manufacturing plant has a maximum allowable airborne particulate limit of

0.100 ppm within the plant clean room. If the clean room contains a total of

450. kg of air, what would be the maximum allowed mass (in grams) of total

airborne particulates allowed in the clean room?



A property of a solution that is dependent on concentration is known as a

Colligative Property.

Colligative Properties of Solutions: (Electrolytes)

An ionic compound or acid that forms an aqueous solution that

conducts electricity is an electrolyte. Electrolytes have free-moving

ions in solution that allow the electrons to flow through the solution.

Pure (distilled) water will not flow electricity as it contains no

dissolved ions (I personally wouldn’t work with electricity while

standing in a puddle containing distilled water!)

Electrolytes will 100% ionize in water in a reaction that resembles a

decomposition reaction. The ionization reaction is called

dissociation, and is a physical change, not a chemical change. The

more ions a solution forms when it dissociates, the higher the boiling

point and the lower the freezing point of the electrolytic solution.

Topic: Colligative Properties

Objective: What properties are based on the solution concentration?



Colligative Properties Examples: (Electrolytes)

1. NaCl(s) Na+1

(aq) + Cl-1

(aq)

a. One mole of sodium chloride dissociates into one mole of sodium

cations and one mole of chlorine anions (two total moles of ions).

2. CaBr2(s) Ca+2

(aq) + 2 Br-1

(aq)

a. One mole of calcium bromide dissociates into one mole of calcium

cations and two moles of chlorine anions (three total moles of ions).

3. Al(NO3)3(s) Al+3

(aq) + 3 NO3-1

(aq)

a. One mole of aluminum nitrate dissociates into one mole of

aluminum cations and three moles of polyatomic nitrate anions (four

total moles of ions).



Nonelectrolytes:

Covalently bonded substances (except many acids) do not become

electrolytes when forming aqueous solutions. This includes polar

molecules that dissolve, but not dissociate, in water.

Examples of polar molecules that dissolve but don’t dissociate are

sugars (C6H12O6, C12H22O11), ethylene glycol (CH2OHCH2OH), and

ethanol (C2H5OH). These solutes have less of an effect on the

freezing and boiling points of solutions than do ionic compounds as

they don’t dissociate further.

Nonelectrolyte examples:

1. C12H22O11(s) C12H22O11(aq)

a. One mole of solid sucrose dissolves in water to form one mole of

aqueous sucrose solution. No ions are formed, and no charges exist

to flow electrons.

2. CH2OHCH2OH(l) CH2OHCH2OH(aq)

a. One mole of ethylene glycol dissolves in water to form one mole of

aqueous ethylene glycol solution. No ions are formed, and no

charges exist to flow electrons.



Colligative Properties of Solutions: (Phase change)

1. Freezing Point Depression:

The freezing point temperature of water is depressed (decreased)

when a solute is added to the water.

2. Boiling Point Elevation:

The boiling point temperature of water elevates (increases) when

a solute is added to the water.

3. The higher the concentration of a solute in a solution, the lower the

freezing point (larger depression) and the higher the boiling point

(larger elevation) for the solution.

4. The more particles that a solute dissolves or dissociates into, the

lower the freezing point (larger depression) and the higher the boiling

point (larger elevation) for the solution.

Watch Crash Course Chemistry Water and Solutions video 13:33

Topic: Colligative Properties

Objective: What properties are based on the solution concentration?

https://www.youtube.com/watch?v=AN4KifV12DA&index=7&list=PL8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr



Colligative Properties Examples: (Phase Change)

1. Which of the following aqueous solutions would boil at the highest

temperature?

a) 100 g of NaCl in 125 g of water

b) 100 g of NaCl in 250 g of water

c) 100 g of NaCl in 500 g of water

d) 100 g of NaCl in 1000 g of water

Concentration = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡

i. If the amount of solute (NaCl) remains the same, the less solvent (water)

there is would make the solution more concentrated.

ii. As concentration increases, the boiling point elevates. The answer

would then be choice ‘A’.

2. Which of the following aqueous solutions would boil at the lowest temperature?





i. Concentration = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒


ii. If the amount of solute (NaCl) remains the same, the more solvent (water)

there is would make the solution less concentrated.

iii. As concentration decreases, the boiling point decreases. The answer would

then be choice ‘D’.

3. Which of the following aqueous solutions would freeze at the lowest

temperature?









ii. If the amount of solute (NaCl) remains the same, the less solvent (water) there

is would make the solution more concentrated.

iii. As concentration increases, the freezing point depresses. The answer would

then be choice ‘A’.

4. Which of the following aqueous solutions would freeze at the highest

temperature?







ii. If the amount of solute (NaCl) remains the same, the more solvent (water)

there is would make the solution less concentrated.

iii. As concentration increases, the freezing point increases. The answer would

then be choice ‘D’.

5. Which of the following aqueous solutions will boil at the highest temperature?

a) 1 mole of KBr in 500 g of water

b) 1 mole of MgF2 in 500 g of water

c) 1 mole of AlCl3 in 500 g of water

d) 1 mole of C6H12O6 in 500 g of water

i. The ionic compound AlCl3 has a total of four atoms, and each dissociates into

a separate ion, so a single mole of AlCl3 makes four moles of ions. The more

particles that a solute forms, the higher the boiling point elevates, so the

answer is ‘C’.



6. Which of the following aqueous solutions will boil at the lowest temperature?





i. The ionic compound C6H12O6 has a total of 1 molecule, so a single mole of

C6H12O6 makes one mole of particles. The fewer particles that a solute forms,

the lower the boiling point elevates, so the answer is ‘D’.

7. Which of the following aqueous solutions will freeze at the lowest temperature?





i. The ionic compound AlCl3 has a total of four atoms, and each dissociates into

a separate ion, so a single mole of AlCl3 makes four moles of ions. The more

particles that a solute forms, the lower the freezing point depresses, so the

answer is ‘C’.

8. Which of the following aqueous solutions will freeze at the highest

temperature?





i. The ionic compound C6H12O6 has a total of 1 molecule, so a single mole of

C6H12O6 makes one mole of particles. The fewer particles that a solute forms,

the less is the effect on the freezing point, so the answer is ‘D’.



Colligative Properties Regents Practice Problems (ungraded):

1. Which aqueous solution will have the lowest freezing point?

a) 1.0 M NaCl

b) 1.0 M C6H12O6

c) 1.0 M C2H5OH

d) 1.0 M CH3COOH

2. As a solute is added to a solvent, what happens to the freezing point

and the boiling point of the solution?

a) The freezing point decreases and the boiling point decreases.

b) The freezing point decreases and the boiling point increases.

c) The freezing point increases and the boiling point decreases.

d) The freezing point increases and the boiling point increases.

3. Compared to pure water, an aqueous solution of calcium chloride has

a

a) Lower boiling point and a lower freezing point.

b) Lower boiling point and a lower freezing point.

c) Higher boiling point and a lower freezing point.

d) Lower boiling point and a higher freezing point.

4. A student dissolves 1.0 mole of sucrose (C12H22O11) in 1000. grams

of water at 1.0 atmosphere. Compared to the boiling point of pure

water, the boiling point of the resulting solution is

1. 273 K

2. 372 K

3. 373 K

4. 374 K

5. Which 1.0 M aqueous solution would have the lowest freezing point?



The chart above is used to determine the effect that adding

ethylene glycol (antifreeze) to your cooling system will have

given the concentration.

Using the chart above, determine:

6. How many quarts of antifreeze are needed to protect a 16-quart

cooling system down to -28F?

7. Based on the bottommost (FREEZE/BOIL PROTECTION CHART)

chart, will aqueous ethylene glycol provide a colligative property

effect against freezing and/or boiling? Explain your answer.



Notes page:





Colligative Properties Homework

Ionic compounds (containing BOTH a metal and a nonmetal) are broken apart by water

molecules into individual ions. Only ionic bonds may be broken by dissolving, and any

covalent bonds within a polyatomic ion (Reference Table E) remain intact.

For each of the following compounds determine which ions make up the compound and

how many total ions are contained within the compound. 1 pt. ea.

Ionic compound # of and symbol of

cation

# of an symbol of

anion

Total # of ions in

compound

KHCO3

Fe(C2H3O2)2

Mg(OH)2

CaBr2

Ionic compounds (as described above) form ions, so the resulting aqueous solutions are

known as electrolytes. Covalent molecules (containing only nonmetals) do not form ions,

and therefore do not act as electrolytes. However, for ionic compounds and covalent

compounds, the greater the number of separate aqueous molecules (concentration) in a

solution will have a greater impact on both freezing and boiling point. 1 pt. ea.

Compound Ionic or

Molecular?

Electrolyte of

Nonelectrolyte?

How many

aqueous

particles are

formed per?

Rank the order of

freezing/boiling point

impact (1-less impact;

4-most impact)

BaBr2

LiF

C2H6O

Fe(NO3)3

Cont’d next page



Multiple choice questions: Circle the correct answer. 1 pt. ea.

1. Which of the following 1 M solutions will have the highest boiling point?

a) NaCl(aq)

b) C6H12O6(aq)

c) K3PO4(aq)

d) Cu(NO3)2(aq)

2. Which of the following 1 M solutions will have the lowest boiling point?

a) NaCl(aq)

b) C6H12O6(aq)

c) K3PO4(aq)

d) Cu(NO3)2(aq)

3. Which of the following 1 M solutions will have the highest freezing point?

a) NaCl(aq)

b) C6H12O6(aq)

c) K3PO4(aq)

d) Cu(NO3)2(aq)

4. Which of the following 1 M solutions will have the lowest freezing point?

a) NaCl(aq)

b) C6H12O6(aq)

c) K3PO4(aq)

d) Cu(NO3)2(aq)

5. Which of the following aqueous solutions will have the highest boiling point?

a) 500. g of solute in 1.00 kg of solvent

b) 500. g of solute in 0.500 kg of solvent

c) 1000. g of solute in 1.00 kg of solvent

d) 1000. g of solute in 0.500 kg of solvent

6. Which of the following aqueous solutions will have the lowest boiling point?





7. Which of the following aqueous solutions will have the highest freezing point?





8. Which of the following aqueous solutions will have the lowest freezing point?







Notes page:

unit 10: solutions-lecture regents chemistry ’14 mr ... · unit 10: solutions-lecture regents...

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