Solution
Stoichiometry
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
1
Measures of Concentration
The concentration of a solution is the amount of solute present in a
given quantity of solvent or solution.
Molarity (M)
moles of solute
M =
liters of solution
Includes solute
volume
Because density (volume) can change with temperature it is helpful to
express solvent by mass when sample undergoes temperature changes
Molality (m)
m =
moles of solute
mass of solvent (kg)
Excludes solute
mass
No volumetric measurements needed; all mass
(grams)
M =
moles of solute
liters of solution
(mL)
moles = Liters x Molarity
L =
moles of solute
Molarity
3
Molarity/Molality Problem
Calculate the Molarity and Molality of a H2SO4
solution containing 24.4 g of sulfuric acid in 198 g
of water at 70°C to produce a 204 mL solution.
M =
moles
m =
L solution
moles
kg solvent
24.4 g H2SO4 1 mol H2SO4
= 0.249 mol H2SO4
98.1 g H2SO4
M =
0.249 mol
0.204 L
= 1.22 Molar H2SO4
m =
0.249 mol
0.198 kg
= 1.26 molal H2SO4
Ion Molarity Problem
What is the Molar concentration of the Potassium ion [K+]
when 9.85 g of K2CO3 are dissolved in a 250 mL solution?
9.85 g K2CO3 1 mol K2CO3
2 mol K+
138.2 g K2CO3 1 mol K2CO3
K2CO3(s) → 2K+(aq) + CO3-2(aq)
M =
0.143 mol
0.250 L
= 0.143 mol K+
= 0.57 M K+
5
Preparing a Solution of Known Concentration from solids
Volumetric Flask
Mix till dissolved
Bring to
desired volume
Molarity Problem
What is the Molar concentration of the Sodium ion [Na+] when
23.4 g of NaCl and 34.1 g of Na2O are dissolved in 0.60 L H2O?
23.4 g NaCl 1 mol
58.5 g NaCl
34.1 g Na2O 1 mol Na2O
= 0.40 mol NaCl = 0.40 mol Na+
2 mol Na+
62.8 g Na2O 1 mol Na2O
= 1.08 mol Na+
(doubles from subscript)
0.40 mol Na+ + 1.08 mol Na+ = 1.48 mol Na+
M =
1.48 mol
0.60 L
= 2.5 M Na+
7
Molarity Problem
How many grams of potassium dichromate (K2Cr2O7)
are required to prepare a 250-mL solution whose
concentration is 2.16 M?
M =
mol
L
mol = ML
Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution (stock).
Dilution
Add Solvent
Moles of solute
before dilution (1)
M1V1
=
=
Moles of solute
after dilution (2)
M2V2
Number of moles does not change
9
4.9 Dilution Practice
Describe how you would prepare 500 mL
of a 1.75 M H2SO4 solution, starting with
an 8.61 M stock solution of H2SO4.
Keep in mind that in dilution, the concentration of the solution
decreases but the number of moles of the solute remains the same.
M1V1 = M2V2
4.9 Solution
Solution We prepare for the calculation by tabulating our data:
M1 = 8.61 M
V1 = ?
M2 = 1.75 M
V2 = 500 mL
Thus, we must dilute 102 mL of the 8.61 M H2SO4 solution with
water to give a final volume of 500 mL
Dilution Problem
Describe how you would prepare 300
mL of a 0.4 M H3PO4 solution, starting
with an 1.5 M stock solution of H3PO4.
M1V1 = M2V2
Bell Ringer
You have 250 mL of a 3.0 M Ba(OH)2 solution.
What is the concentration if we add 150 mL of
water to the solution?
M1 = 3.0 M
V1 = 250 mL
M2 = ?
V2 = 250 mL + 150 mL = 400 mL
Crash Course: Water and Solutions for Dirty Laundry
www.youtube.com/watch?v=AN4KifV12DA
Bell Ringer
1) How many grams of solid NaNO3 are needed to produce
125 mL of a 0.85 M NaNO3 solution?
2) What is the Molar concentration of the Sodium ion [Na+]
when 2.8 g of Na3PO4 and 4.5 g of Na2CO3 are dissolved in
85 mL?
3) How would you prepare 250 mL of a 0.65 M H2SO4
solution from a stock of 6.5 M H2SO4 solution?
4) You have 50 mL of 6.0 M NaF and 450 mL water are
added, what is the new Molarity?
14
Titrations
In a titration, a solution of accurately known concentration is
added gradually to another solution of unknown concentration
until the chemical reaction between the two solutions is
complete.
Standard solution – solution with known concentration to be
precisely added for comparison
Equivalence point – the point at which the reaction is complete
example) 1 mol H2SO4  2 mol NaOH
(obtained from balanced equation)
15
Titrations
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
Titrations can be used in the analysis of:
Acid-base reactions
H2SO4 + 2NaOH
2H2O + Na2SO4
Redox reactions
5Fe2+ + MnO4- + 8H+
Mn2+ + 5Fe3+ + 4H2O
*Can involve color changes without indicator
17
Titration Steps
1) Write Balanced equation for stoichiometry
(mole-to-mole ratio)
2) Determine moles of Standard solution used.
(mol = M x L)
3) With stoichiometry, convert molesstandard to molesunknown
(train tracks)
4) Determine unknown Molarity using given volume (L)
(M = mol/L)
Alternative Equation:
MsVs = MuVu
Coefficient #
Coefficient #
Titration Problem #1
It takes 32 mL of 2.0M HCl standard to neutralize a 500. mL
solution of Ba(OH)2. What is the concentration of Ba(OH)2?
1) 1Ba(OH)2 + 2HCl → BaCl2 + 2H2O (Reacts 1:2)
2) mol HCl = 2.0M x 0.032 L = 0.064 mol HCl
3) 0.064 mol HCl 1 mol Ba(OH)2
2 mol HCl
4) M Ba(OH)2 = 0.032 moles
0.500 L
= 0.032 mol Ba(OH)2
= 0.064 M Ba(OH)2
Alternative:
MHVH = MOHVOH
Coefficient #
Coefficient #
2.032 = MOH500
2
1
MOH = 0.064
Titration Problem #2
How many milliliters (mL) of a 0.610 M NaOH
solution are needed to neutralize 20.0 mL of a
0.245 M H2SO4 solution?
2 NaOH + H2SO4
2 H2O + Na2SO4
1) For every 2 moles base added,
it neutralizes 1 mole acid
Titration #2 Solution
2) Next we calculate the number of moles of H2SO4 in a
20.0 mL solution:
Moles = M x L
0.245 M x 0.0200 L = 0.00490 mol H2SO4
3) From the Balanced Equation: 1 mol H2SO4  2 mol NaOH.
4.9 x 10-3 mol H2SO4
2 mol NaOH
= 9.80 × 10-3 mol
1 mol H2SO4
NaOH
4) L = 9.80 x 10-3 mol = 0.0161 L or 16.1 mL NaOH
0.61 M NaOH
Titration: Finding the Molar Mass of an Unknown
Lauric Acid is a short-chain fatty acid that is solid at room
temperature and monoprotic. We dissolve 0.022 grams into
500. mL of water and then titrate it with 0.010 M NaOH.
If it takes 11.0 mL of NaOH to neutralize the fatty acid, what
is the Molar Mass of Lauric Acid?
• Molar mass has the units grams/mole. We weighed out
the mass of the solid acid in grams. Titration can tell us
how many moles of acid are present in the same sample.
0.022 grams
= ? Molar mass
+
Moles H
• Because it is monoprotic, it will react 1:1 with NaOH.
Molar Mass Titration Solution
1) Given information states it reacts 1:1
2) (0.010 M NaOH) x (0.0110 L) = 1.1 x 10-4 mol NaOH
3) 1.1 x 10-4 mol NaOH 1 mol Lauric acid = 1.1 x 10-4 mol
Lauric Acid
1 mol NaOH
0.022 grams
4) Molar Mass =
1.1 x 10-4 moles
C11H23COOH
= 200 g/mol
Large component of
coconut oil
~ 3-6% of milk
Example: Redox Titration
A 16.42-mL volume of 0.1327 M KMnO4
solution is needed to oxidize 25.00 mL of a
FeSO4 solution in an acidic medium.
What is the concentration of the FeSO4
solution in molarity?
The net ionic equation is
need to
find
want to
calculate
given
Redox Titration Solution
Solution The number of moles of KMnO4 (in 16.42 mL) = M x L
(0.1327 M KMnO4) x (0.01642 L) = 2.179 x 10-3 mol
From the net ionic equation we see that 5 mol Fe2+  1 mol MnO4-
M =
moles of solute
liters of solution
1.090 x 10-2 mol
=
0.025 L
= 0.436 M FeSO4
Titration Bell Ringer
• 50 mL of Ba(OH)2 is measured out into an Erlenmeyer
flask. The concentration is unknown.
• It takes 8.5 mL of 2.5 M H3PO4 standard to reach the
equivalence point and neutralize the unknown base.
• Determine the concentration of Ba(OH)2 solution.
26
Chemistry in Action: Metals from the Sea
Many metals are found
in the earth’s crust,
but it is cheaper to
“mine” from the sea
CaCO3 (s)
CaO (s) + CO2 (g)
Precipitation
Ca2+ (aq) + 2OH- (aq)
CaO (s) + H2O (l)
Slightly soluble
Mg2+ (aq) + 2OH- (aq)
Mg(OH)2 (s)
Precipitation
Mg(OH)2 (s) + 2HCl (aq)
MgCl2 (aq) + 2H2O (l)
1.3 g of Magnesium/
Kg seawater
Electrolysis of MgCl2 (redox)
Mg2+ + 2e-
2ClMgCl2 (aq)
Mg
Cl2 + 2eMg (s) + Cl2 (g)
27
Gravimetric Analysis
 Analytical technique based on the measurement of mass
• Precipitation: the analyte is precipitated out of solution by adding
another reagent to make it insoluble. Then filtered and weighed.
• Volatilization: the analyte is converted to a gas and removed. The
loss of mass from the starting material indicates the mass of gas.
28
Gravimetric Analysis
1. Dissolve unknown substance in water (if not already dissolved)
2. React unknown with precipitating reagent to form a solid precipitate
Reagent: chemical added to another substance to bring about a change
3. Filter, dry, and weigh precipitate.
4. Use chemical formula and mass of precipitate to determine amount of
unknown analyte (chemical of interest in experiment)
29
Gravimetric Analysis Problem #1
A sample of an unknown soluble compound contains Ba+2 and is
dissolved in water and treated with excess sodium phosphate.
If 0.411 g of Barium phosphate precipitates out of solution, what
mass of Barium in found in the unknown compound?
We need to find 1st find the Mass % of the analyte in the precipitated
compound using their respective molar masses
3 x 137.3 g Ba+2
601.9 g Ba3(PO4)2
x 100% = 68.4% Ba in Ba3(PO4)2
0.684 x 0.411 g = 0.253 g Ba+2
*It is not mandatory to convert to
percentage form. The mass
fraction can be used directly.
411.9 g Ba+2
+2
x
0.411g
Ba
(PO
)
=
0.253
g
Ba
3
4 2
601.9 g Ba3(PO4)2
Gravimetric Analysis Problem #2
We have 250 mL of a Copper (Cu+1) solution. We add excess
Na2CO3 to precipitate out 3.8 g of Cu2CO3. What is the [Cu+1]
Molarity of the original solution?
Since we need to find moles of Cu+1 , it will be quicker to use train-tracks
instead of % composition (either would work).
3.8 g Cu2CO3 1 mol Cu2CO3 2 mol Cu+1
187.0 g Cu2CO3 1 mol Cu2CO3
M =
0.041 mol Cu+
0.250 L
= 0.16 M Cu+1
= 0.041 mole Cu+1
Gravimetric Analysis Problem #3
A 0.5662-g sample of an ionic compound containing
chloride ions and an unknown metal is dissolved in water
and treated with an excess of AgNO3.
If 1.0882 g of AgCl precipitate forms, what is the percent
by mass of Cl in the original compound?
35.45 g Clx 1.0882 g AgCl = 0.269 grams Cl
143.4 g AgCl
More Gravimetric Review Problems
1) An unknown ionic compound contains Carbonate (CO3-2).
To precipitate the carbonate, we add excess CaCl2 and collect
25.3 grams of CaCO3 precipitate. Calculate mass of Carbonate
present in the original compound.
2) 50 mL of a solution contains an unknown amount of Ni+
ions. We add excess Na3PO4 to precipitate out 5.67 grams
of Ni3PO4. What is the Molarity of Ni+?
3) 340 mL of an unknown solution contains the Silver ion
(Ag+). When excess Na2S is added, 15.8 grams of
precipitated Ag2S are formed. What is the Molarity of
Ag+ in the solution?