Molarity, Dilution, and pH Main Idea: Solution concentrations are measured in molarity. Dilution is a useful technique for creating a new solution from

Molarity, Dilution, and pH

Main Idea: Solution concentrations are measured in molarity. Dilution is a useful technique for creating a new solution from a stock solution. pH is a measure of the concentration of hydronium ions in a solution.

1

Properties of Aqueous Solutions

• Solution- a homogeneous mixture of two or more substances.

• Solute- a substance in a solution that is present in the smallest amount.

• Solvent- a substance in a solution that is present in the largest amount.

• In an aqueous solution, the solute is a liquid or solid and the solvent is always water.

Molarity Review• One of the most common units of solution

concentration is molarity.• Molarity (M) is the number of moles of solute per

liter of solution.• Molarity is also known as molar concentration,

and the unit M is read as “molar.”• A liter of solution containing 1 mol of solute is a

1M solution, which is read as a “one-molar” solution.

• A liter of solution containing 0.1 mol of solute is a 0.1 M solution.

3

Molarity Equation

• To calculate a solution’s molarity, you must know the volume of the solution in liters and the amount of dissolved solute in moles.

• Molarity (M) = moles of soluteliters of solution

4

Molarity ExampleA 100.5-mL intravenous (IV) solution contains 5.10 g of

glucose (C6H12O6). What is the molarity of the solution? The molar mass of glucose is 180.16 g/mol.

SOLUTION:1) Calculate the number of moles of C6H12O6 by dividing

mass over molar mass = 0.0283 mol C6H12O6 2) Convert the volume of H2O to liters by dividing volume

by 1000 = 0.1005 L3) Solve for molarity by dividing moles by liters = 0.282 M

5

Preparing Molar Solutions• Now that you know how to calculate the molarity of a

solution, how would you prepare one in the laboratory?

• STEP 1: Calculate the mass of the solute needed using the molarity definition and accounting for the desired concentration and volume.

• STEP 2: The mass of the solute is measured on a balance.

• STEP 3: The solute is placed in a volumetric flask of the correct volume.

• STEP 4: Distilled water is added to the flask to bring the solution level up to the calibration mark.

6

http://www.ltcconline.net/stevenson/2008CHM101Fall/CHM101LectureNotes20081022.htm

7

Properties of Aqueous Solutions

• All solutes that dissolve in water fit into one of two categories: electrolyte or non-electrolyte.

• Electrolyte- a substance that when dissolved in water conducts electricity

• Non-electrolyte- a substance that when dissolved in water does not conduct electricity.

• To have an electrolyte, ions must be present in water.

Electrolytic Properties of Aqueous Solutions

• NaCl in water.– What happens?– NaCl(s) → Na+(aq) + Cl–(aq) – Completely dissociates

Strong vs. Weak Electrolytes

• How do you know when an electrolyte is strong or weak?

• Take a look at how HCl dissociates in water.– HCl(s) → H+(aq) + Cl–(aq)



Hydrated Ions


• What about weak electrolytes? • What makes them weak?

– Ionization of acetic acid

• CH3COOH(aq) ↔ CH3COO–(aq) + H+(aq)

Electrolytic Solutions

Precipitation Reactions

• Precipitation Reaction- a reaction that results in the formation of an insoluble product.

• These reactions usually involve ionic compounds.

• Formation of PbI2:– Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)

Preciptate

Precipitate


• How do you know whether or not a precipitate will form when a compound is added to a solution?

• By knowing the solubility of the solute!• Solubility- The maximum amount of solute that will

dissolve in a given quantity of solvent at a specific temperature.

• Three levels of solubility: Soluble, slightly soluble or insoluble.


Determining Solubility

• Determine the solubility for the following: (1) Ag2SO4

(2) CaCO3

(3) Na3PO4

Diluting Molar Solutions• In the laboratory, you might use concentrated

solutions of standard molarities, called stock solutions.– For example, concentrated hydrochloric acid (HCl) is

12 M.• You can prepare a less-concentrated solution by

diluting the stock solution with additional solvent.– Dilution is used when a specific concentration is

needed and the starting material is already in the form of a solution (i.e., acids).

22

Dilution of Solutions

• When you want to dilute a solution, what happens to the number of moles present in the solution?– Do they increase?– Decrease?– Stay the same?

Dilution of Solutions

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?

Add water to the 3.0 M solution to lower its Add water to the 3.0 M solution to lower its concentration to 0.50 M concentration to 0.50 M

Dilute the solution!Dilute the solution!

25

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you do?do?


3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how much water But how much water do we add?do we add?

26

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you dodo??

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you dodo??

How much water is added?How much water is added?The important point is that --->The important point is that --->

moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution

27

PROBLEM: You have 50.0 mL of 3.0 M NaOH and PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOH and PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?you want 0.50 M NaOH. What do you do?

Amount of NaOH in original solution = Amount of NaOH in original solution = M • V = = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH(3.0 mol/L)(0.050 L) = 0.15 mol NaOHAmount of NaOH in final solution must also

= 0.15 mol NaOHVolume of final solution =Volume of final solution =(0.15 mol NaOH) / (0.50 M) = 0.30 Lor or 300 mL

28



Conclusion:Conclusion:

add 250 mL of add 250 mL of waterwater to 50.0 to 50.0 mL of 3.0 M mL of 3.0 M NaOH to make NaOH to make 300 mL of 0.50 300 mL of 0.50 M NaOH. M NaOH. 3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute29

A shortcutA shortcut

MM11 • V • V11 = M = M22 • V • V22

Where M represents molarity and V Where M represents molarity and V represents volume. The 1s are for represents volume. The 1s are for the stock solution and the 2s are for the stock solution and the 2s are for the solution you are trying to create.the solution you are trying to create.

Preparing Solutions by Preparing Solutions by DilutionDilution

Preparing Solutions by Preparing Solutions by DilutionDilution

30

What factors affect the rate of dissolving?

31


• Temperature• You can dissolve more into a warm liquid than you can

into a cold liquid

32



into a cold liquid

• Surface area• Which dissolves faster, a cube of sugar or grains of

sugar? Why?

33



into a cold liquid

• Surface area• Which dissolves faster, a cube of sugar or grains of

sugar? Why?

• Concentration• The more solute already dissolved in a solvent, the

slower the rate of dissolving.

34


• Pressure• What affect would increasing pressure have on the rate

of dissolving? Why?

35


• Pressure• What affect would increasing pressure have on the rate

of dissolving? Why?

• Mixing• Describe what happens when you mix a solution.

36

How do you get from this…

…to this?

Add an ionic compound!

Colligative Properties

• Properties that depend only on the number of solute particles and not on their identity.

Some Colligative Properties are:

• Vapor pressure lowering• Boiling point elevation• Freezing Point depression

Vapor Pressure

Vapor Pressure Lowering

• The particles of solute are surrounded by and attracted to particles of solvent.

• Now the solvent particles have less kinetic energy and tend less to escape into the space above the liquid.

• So the vapor pressure is less.

Ionic vs Molecular Solutes

• Ionic solutes produce two or more ion particles in solution.

• They affect the colligative properties proportionately more than molecular solutes (that do not ionize).

• The effect is proportional to the number of particles of the solute in the solution.

How many particles do each of the following give upon solvation?

• NaCl• CaCl2

• Glucose

Freezing Point Depression

Example

• Salt is added to melt ice by reducing the freezing point of water.

Boiling Point Elevation

Example• Addition of ethylene glycol C2H6O2 (antifreeze)

to car radiators.

Ready for a test soon?!?!?!

• It’s review time!• In this unit we studied SOLUTIONS!

50

Freezing Point Depression and Boiling Point Elevation

Boiling Point Elevation • ∆Tb =mkb (for water kb=0.51 oC/m)

• Freezing Point Depression• ∆Tf=mkf (for water kf=1.86 oC/m)

• Note: m is the molality of the particles, so if the solute is ionic, multiply by the #of particles it dissociates to.

Which is more effective for lowering the freezing point of water?

• NaCl or CaCl2

Example 1:

• Find the new freezing point of 3m NaCl in water.

Example 2:

• Find the new boiling point of 3m NaCl in water.

TheThe pH scalepH scale is a way of is a way of expressing the strength of expressing the strength of acids and bases. Instead of acids and bases. Instead of using very small numbers, using very small numbers, we just use the NEGATIVE we just use the NEGATIVE power of 10 on the Molarity power of 10 on the Molarity of the Hof the H++ (or OH (or OH--) ion.) ion.

Under 7 = acidUnder 7 = acid 7 = neutral 7 = neutral

Over 7 = baseOver 7 = base55

Acid-Base Reactions

• Acids- generally have a sour taste, change litmus from blue to red, can react with certain metals to produce gas, conduct electricity.

• Bases- generally have a bitter taste, change litmus from red to blue, feel slippery, conduct electricity.

• BrØnstead Acid- proton donor• BrØnstead Base- proton acceptor

Acid-Base Reactions

• Acid or Base?– HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)

– NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq)

Acid-Base Reactions

• Look at the following compounds and decide whether they are a BrØnstead Acid or a BrØnstead Base.– HBr– NO2

–

– HCO3–

59

Arrhenius definition of acids and bases: Acids are compounds that give off H+

ions (also called hydronium ions, H3O+ ions, or simply “protons”) when you dissolve them in water.

Bases are compounds that give off OH- (hydroxide) ions when you dissolve them in water.

• Arrhenius acids almost always start with the letter “H” in their formulas – this is the source of the H+ ion that comes off when you dissolve the compound. Common examples include the following:

• HNO3(l) H+(aq) + NO3(aq)

-

• HCl• HBr• H2SO4

• Arrhenius bases always have “OH” in their formulas, indicating the presence of the hydroxide ion. Common examples include:

• NaOH(s) Na+1(aq) + OH-1

(aq)

• KOH• Mg(OH)2 60

Acid-Base Reactions

pH of Common SubstancespH of Common Substances

62

pH Value H+ ConcentrationRelative to Pure Water Example

0 10 000 000 battery acid

1 1 000 000 concentrated sulfuric acid

2 100 000 lemon juice, vinegar

3 10 000 orange juice, soda

4 1 000 tomato juice, acid rain

5 100 black coffee, bananas

6 10 urine, milk

7 1 pure water

8 0.1 sea water, eggs

9 0.01 baking soda

10 0.001 Great Salt Lake, milk of magnesia

11 0.000 1 ammonia solution

12 0.000 01 soapy water

13 0.000 001 bleach, oven cleaner

14 0.000 000 1 liquid drain cleaner 63

Put the following substances in order, from lowest pH to highest pH:

• Drain Cleaner• Water• Vinegar• Soap• Orange Juice• Baking Soda• Battery Acid

64

Put the following substances in order, from lowest pH to highest pH:

• Battery Acid Lowest pH (most Acidic)• Vinegar• Orange Juice• Water• Baking Soda• Soap• Drain Cleaner Highest pH (most Basic)

65

Calculating the pH

pH = - log [H+](The [ ] means Molarity)

Example: If [H+] = 1 X 10-10

pH = - log 1 X 10-10

pH = - (- 10)pH = 10

Example: If [H+] = 1.8 X 10-5

pH = - log 1.8 X 10-5

pH = - (- 4.74)pH = 4.74

66

pH calculations – Solving for pH calculations – Solving for H+H+pH calculations – Solving for pH calculations – Solving for H+H+

If the pH of Coke is 3.12, [HIf the pH of Coke is 3.12, [H++] = ???] = ???

Because pH = - log [HBecause pH = - log [H++] then] then

- pH = log [H- pH = log [H++]]

Take antilog (10Take antilog (10xx) of both) of both sides and get sides and get

1010-pH -pH == [H[H++]][H+] = 10-3.12 = 7.6 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd function” and

then the log button67

pH calculations – Solving for H+pH calculations – Solving for H+• A solution has a pH of 8.5. What is the A solution has a pH of 8.5. What is the

Molarity of hydrogen ions in the solution?Molarity of hydrogen ions in the solution?

pH = - log [HpH = - log [H++]]

8.5 = - log [H8.5 = - log [H++]]

-8.5 = log [H-8.5 = log [H++]]

Antilog -8.5 = antilog (log [HAntilog -8.5 = antilog (log [H++])])

1010-8.5-8.5 = [H = [H++]]

3.16 X 103.16 X 10-9-9 = [H = [H++]]

pH = - log [HpH = - log [H++]]

8.5 = - log [H8.5 = - log [H++]]

-8.5 = log [H-8.5 = log [H++]]

Antilog -8.5 = antilog (log [HAntilog -8.5 = antilog (log [H++])])

1010-8.5-8.5 = [H = [H++]]

3.16 X 103.16 X 10-9-9 = [H = [H++]]68

More About WaterMore About WaterMore About WaterMore About WaterHH22O can function as both an ACID and a O can function as both an ACID and a

BASE…..water is BASE…..water is AMPHOTERICAMPHOTERIC

In pure water there can be In pure water there can be AUTOIONIZATIONAUTOIONIZATION

Equilibrium constant for water = KEquilibrium constant for water = Kww

KKww = [H = [H33OO++] [OH] [OH--] =] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC

More About WaterMore About Water

KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC

In a In a neutral neutral solution [Hsolution [H33OO++] = [OH] = [OH--]]

so Kso Kww = [H = [H33OO++]]22 = [OH = [OH--]]22

and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M

OH-

H3O+

OH-

H3O+

AutoionizationAutoionization

70

pOH• Since acids and bases are opposites, pH Since acids and bases are opposites, pH

and pOH are opposites!and pOH are opposites!• pOH does not really exist, but it is pOH does not really exist, but it is

useful for changing bases to pH.useful for changing bases to pH.• pOH looks at the perspective of a basepOH looks at the perspective of a base

pOH = - log [OHpOH = - log [OH--]]Since pH and pOH are on opposite ends,Since pH and pOH are on opposite ends,

pH + pOH = 14pH + pOH = 14

71

pHpH [H+][H+] [OH-][OH-] pOHpOH72

[H[H33OO++], [OH], [OH--] and pH] and pHWhat is the pH of the What is the pH of the

0.0010 M NaOH solution? 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10[OH-] = 0.0010 (or 1.0 X 10-3-3 M) M) pOH = - log 0.0010pOH = - log 0.0010 pOH = 3pOH = 3pH = 14 – 3 = 11pH = 14 – 3 = 11

OR KOR Kww = [H = [H33OO++] [OH] [OH--]][H[H3OO++] = 1.0 x 10] = 1.0 x 10-11-11 M MpH = - log (1.0 x 10pH = - log (1.0 x 10-11-11) = 11.00) = 11.00

73

[OH[OH--]]

[H[H++]] pOHpOH

pHpH

1010 -pOH-pOH

1010 -pH-pH-Log[H

-Log[H ++]]

-Log[OH

Log[OH --]]

14 - pOH

14 - pOH

14 - pH

14 - pH

1.0 x 10

1.0 x 10-

14-14

[OH[OH

-- ]]

1.0 x 10

1.0 x 10-

14-14

[H[H++ ]]

74

Oxidation Reduction Reactions

• Oxidation Reaction- refers to the half-reaction that involves the loss of electrons.

• Reduction Reaction- refers to the half-reaction that involves the gain of electrons.

OILRIG• Oxidizing agent- the compound or ion in a redox

reaction that donates electrons.• Reducing agent- the compound or ion in a redox

reaction that accepts electrons.

Oxidation-Reduction Reactions

HOMEWORK

1) How much calcium hydroxide [Ca(OH)2], in grams, is needed to produce 1.5 L of a 0.25 M solution?

2) What volume of a 3.00M KI stock solution would you use to make 0.300 L of a 1.25 M KI solution?

3) How many mL of a 5.0 M H2SO4 stock solution would you need to prepare 100.0 mL of 0.25 M H2SO4?

4) If 0.50 L of 5.00 M stock solution is diluted to make 2.0 L of solution, how much HCl, in grams, is in the solution?

77

HOMEWORK

5) Calculate the pH of solutions having the following ion concentrations at 298 K.a) [H+] = 1.0 x 10-2 M b) [H+] = 3.0 x 10-6 M

6) Calculate the pH of a solution having [OH-] = 8.2 x 10-6 M.

7) Calculate pH and pOH for an aqueous solution containing 1.0 x 10-3 mol of HCl dissolved in 5.0 L of solution.

8) Calculate the [H+] and [OH-] in a sample of seawater with a pOH = 5.60.

78

Documents

Molarity, Dilution, and pH Main Idea: Solution concentrations are measured in molarity. Dilution is a useful technique for creating a new solution from