Concentration of Solutions

Review:

Solutions are made up of

1) Solute - substance dissolved or present in lesser proportion

2) Solvent - substance that is the dissolving medium - what the solute is dissolved in - many times this is water

Solutions can be classified as

1) Electrolyte solutions - solutions that conduct electricity (solute is either an ionic compound or forms ions when it dissolves)

2) Nonelectrolyte solutions - solutions that do not conduct electricity (solute is a molecular compound that does not form ions as it dissolves)

Components that make up mixtures can be

1) miscible - will form solutions in most any proportions (only liquids or gases)

2) immiscible substances - will not form solutions - example: oil and water

There is a limit to the amount of solute that can be dissolved in a given amount of solvent - this is the "solubility" of the solute.

Solubility of a solute can be expressed in the following terms:

1)soluble – The substance mostly dissolves

2)insoluble – Very little of the substance dissolves

3)slightly soluble – In between: some dissolves but it may not be enough to affect the properties of the solution.

Concentration

• The concentration of a solution refers to the amount of solute dissolved in the solvent

• Qualitative terms used are dilute (not much solute) and concentrated (alot of solute) - concentrated does not mean pure.

Molarity

Molarity is one way to measure the concentration of a solution.

moles of solute

volume of solution in litersMolarity (M) =

Molarity is the most used - other units include: molality, normality, formality , mole fraction, % weight, % volume (proof)

Making a Solution…

Sample Problem #1

3.85 g of NaCl is dissolved in enough water to make 81.0 mL of sol’n. Calculate the molarity of the solution.

3.85 g NaCl 1 mol NaCl------------------ X ------------------- = .812M NaCl.0810 L sol’n 58.5 g NaCl

Sample Problem #2

How many grams of NaCl are required to make 450 mL of .500 M soln?

.500 moles NaCl 58.5 g NaCl.45 L soln X ---------------------- X ---------------- = 13 g NaCl 1 L soln 1 mole NaCl

Sample Problem #3

How many mL of a .250 M NaCl soln can be prepared using 7.51 g NaCl?

1 mole NaCl 1 L soln 1000 mL7.51 g NaCl X-------------- X ----------------X -------- = 514mL NaCl soln 58.5 g NaCl .250 mole NaCl 1 L

Titration

Sample Problem #425.0 mL of .325 molar hydrochloric acid (HCl) completely neutralizes 35.0

mL of a calcium hydroxide solution. What is the molarity of the calcium hydroxide solution?

25.0 mL 35.0 mL

.325 M ? M

2 HCl + Ca(OH)2 2 H2O + CaCl2

.0250 L HCl .325 mole HCl 1 mole Ca(OH)2

------------------ X ---------------- X ---------------- = .116 M Ca(OH)2

.0350 L Ca(OH)2 1 L HCl 2 mole HCl

Sample Problem #5How many mL of .525 M nitric acid (HNO3) solution would

completely neutralize 22.5 mL of .275 M Ca(OH)2 base solution?

.525 M .275 M ? mL 22.5 mL2 HNO3 + Ca(OH)2 ----> 2 H2O + Ca(NO3)2

.275 mole Ca(OH)2 2 mole HNO3 1 L HNO3 .0225 L Ca(OH)2 X -------------------- X ------------------- X -----------------

1 L Ca(OH)2 1 mole Ca(OH)2 .525 moles HNO3

= .0236 L or 23.6 mL HNO3

Dilutions

Sample Problem #620.0 mL of .250 M HCl solution is added to 30.0 mL

of .150 M HCl solution. What is the concentration of the resulting solution?

.0200 L X .250 moles/L = .00500 moles HCl

.0300 L X .150 moles/L = .00450 moles HCl ------------ ------------------------- .0500 L .00950 moles HCl total

total molestotal molarity = ------------------

total Liters

.00950 moles HCl M = -------------------------- = .190 M HCl

.0500 L soln.

Sample Problem #710.0 mL of .375 M HCl solution is diluted by

adding water to a new volume of 50.0 mL of solution. What is the concentration of the resulting solution?

total molestotal molarity = -------------------

total Liters

.0100 L X .375 moles/L = .00375 moles HCl

.00375 moles HCl M = ---------------------------- = .0750 M HCl

.0500 L soln.

Sample Problem #7 - 2

10.0 mL of .375 M HCl solution is diluted by adding water to a new volume of 50.0 mL of solution. What is the concentration of the resulting solution?

M1V1 = M2V2

(.375M)(10.0mL) = M(50.0mL)

M = .0750 M

Sample Problem #8Indicate the concentration of each ion present in the solution formed by mixing 44.0 mL of 0.100 M Na2SO4 and 25.0 mL of 0.150 M KCl.

Na2SO4 + KCl X (No reaction!)

Total Volume = 44.0mL + 25.0mL = 69.0mL=.0690L

Sample Problem #91.00 g of aluminum reacts with 75.0 mL of 0.300 M ZnI2

solution. How many grams of zinc are produced? 0.300 M 1.00 g 75.0 mL x g2 Al (s) + 3 ZnI2 (aq) 2 AlI3 (aq) + 3 Zn (s)

0.300 mole ZnI2 2 mole Al 27.0 g Al.0750 L soln x ------------------ x --------------- x ----------- = 0.405 g Al 1 L soln 3 mole ZnI2 1 mole Al

Al is excess or ZnI2 is limiting

0.300 mole ZnI2 3 mole Zn 65.4 g Zn.0750 L soln x ----------------- x ------------- x ------------- = 1.47 g Zn 1 L soln 3 mole ZnI2 1 mole Zn

Sample Problem #1075.0 mL of 0.300 M ZnI2 soln is added to 125 mL of

0.450 M AgNO3 soln. How many grams of the precipitate AgI are produced?

0.300 M 0.450 M75.0 mL 125 mL x gZnI2 (aq) + 2 AgNO3 (aq) 2 AgI (s) + Zn(NO3)2 (aq)

0.300 mole ZnI2 2 mole AgNO3 1 L soln0.0750 L soln x --------------- x --------------- x ------------ = 0.100 L AgNO3

1 L soln 1 mole ZnI2 0.450 mole AgNO3 excess AgNO3 or ZnI2 is limiting

0.300 mole ZnI2 2 mole AgI 235 g AgI0.0750 L soln x ----------------- x --------------- x -------------= 10.6 g AgI 1 L soln 1 mole ZnI2 1 mole AgI