Transcript 4-concentration

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.
Molarity (M) =
moles of solute
volume of solution in liters
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 mL
7.51 g NaCl X-------------- X ----------------X -------- = 514mL NaCl soln
58.5 g NaCl .250 mole NaCl 1 L
Titration
Sample Problem #4
25.0 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 #5
How 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
? mL
2 HNO3
+
.275 M
22.5 mL
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 #6
20.0 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?
total moles
total molarity = -----------------total Liters
.0200 L X .250 moles/L = .00500 moles HCl
.0300 L X .150 moles/L = .00450 moles HCl
-----------------------------------.0500 L
.00950 moles HCl total
.00950 moles HCl
M = -------------------------- = .190 M HCl
.0500 L soln.
Sample Problem #7
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?
total moles
total 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 #8
Indicate 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
0.100 𝑚𝑜𝑙𝑁𝑎2 𝑆𝑂4
2 𝑚𝑜𝑙 𝑁𝑎+
.0440𝐿 𝑁𝑎2 𝑆𝑂4 ∗
∗
= 0.00880 𝑚𝑜𝑙 𝑁
𝐿 𝑁𝑎2 𝑆𝑂4
1 𝑚𝑜𝑙𝑁𝑎2 𝑆𝑂4
.00880𝑚𝑜𝑙
= 0.128 𝑀 𝑁𝑎+
.0690𝐿
0.100 𝑚𝑜𝑙𝑁𝑎2 𝑆𝑂4 1 𝑚𝑜𝑙 𝑆𝑂4 −2
.0440𝐿 𝑁𝑎2 𝑆𝑂4 ∗
∗
= 0.00440 𝑚𝑜𝑙 𝑆𝑂
𝐿 𝑁𝑎2 𝑆𝑂4
1 𝑚𝑜𝑙𝑁𝑎2 𝑆𝑂4
.00440𝑚𝑜𝑙
= 0.0638 𝑀 𝑆𝑂4 −2
.0690𝐿
0.150 𝑚𝑜𝑙 𝐾𝐶𝑙 1 𝑚𝑜𝑙 𝐾 +
.0250𝐿 𝐾𝐶𝑙 ∗
∗
= 0.00375 𝑚𝑜𝑙 𝐾 +
𝐿 𝐾𝐶𝑙
1 𝑚𝑜𝑙 𝐾𝐶𝐿
.00375𝑚𝑜𝑙
= 0.0543 𝑀 𝐾 +
.0690𝐿
0.150 𝑚𝑜𝑙 𝐾𝐶𝑙 1 𝑚𝑜𝑙 𝐶𝑙 −
.0250𝐿 𝐾𝐶𝑙 ∗
∗
= 0.00375 𝑚𝑜𝑙𝐶𝑙 −
𝐿 𝐾𝐶𝑙
1 𝑚𝑜𝑙 𝐾𝐶𝑙
.00375𝑚𝑜𝑙
= 0.0543 𝑀 𝐶𝑙 −
.0690𝐿
Sample Problem #9
1.00 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
xg
2 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 #10
75.0 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 M
75.0 mL
125 mL
xg
ZnI2 (aq) + 2 AgNO3 (aq)  2 AgI (s) + Zn(NO3)2 (aq)
0.300 mole ZnI2 2 mole AgNO3 1 L soln
0.0750 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 AgI
0.0750 L soln x ----------------- x --------------- x -------------= 10.6 g AgI
1 L soln 1 mole ZnI2 1 mole AgI