Molarity

1
Molarity
2. Molarity (M) this is the most common
expression of concentration
M molarity moles of solute
mol liters of solution L
Units are mol L-1 or mol/L (moles per liter)
A solution that is 1.0 molar (1.0M ) contains
1.0 mol of solute per liter of solution
M molarity (mol L-1) n number of moles
(mol) V volume (L)
n M V
M n/V (mol L-1) n M ? V (mol) V n /M (L)
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Example 1 Calculate the molarity of a solution
made by dissolving 11.5g of solid NaOH in enough
water to make 1.50 L of solution.
Step 1 Find the of moles of NaOH
n m 11.5 g MM 40.0g mol-1
0.288 mol
Step2 Find the molarity
M olarity moles of solute 0.288mol liters
of solution 1.50 L
0.192 M NaOH
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Example 2 Calculate the molarity of a solution
made by dissolving 1.56g of gaseous HCl in enough
water to make 26.8 mL of solution.
Step 1 Find the of moles of HCl
n m 1.56 g MM 36.5g mol-1
0.0427 mol HCl 4.27? 10-2 mol HCl
Step2 Convert milliliters to Liters 26.8mL/1000L
0.0268L 2.68 ? 10-2 L
Step3 Find the molarity
M olarity moles of solute 4.27? 10-2
mol liters of solution 2.68 ? 10-2 L
1.59 M HCl
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Example 3 Give the concentrations of all the
ions in each of the following solutions a. 0.50
M Co(NO3)2 b. 1 M FeCl3
a. Co(NO3)2 (s) ?? Co2 (aq) 2NO3- (aq)
1 mol Co(NO3)2 (s) ?? 1 mol Co2 (aq) 2 mol
NO3- (aq)
Therefore a solution that is 0.5 mol Co(NO3)2
contains 0.50 M Co2 and 1 M NO3-
b. FeCl3 (s) ?? Fe3 (aq) 3Cl- (aq)
1 mol FeCl3 (s) ?? 1 mol Fe3 (aq) 3 mol Cl-
(aq)
Therefore a solution that is 1 M FeCl3 contains 1
M Fe3 and 3 M Cl-
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Example 4 How many moles of Ag ions are present
in 25mL of a 0.75 M AgNO3 solution?
Molarity of solution 0.75 M Volume of solution
25mL moles of Ag ions ?
A 0.75 M AgNO3 solution contains 0.75 M Ag ions
and 0.75 M NO3- ions
Convert mL to L 25mL / 1000L 2.5 ? 10-2 L
Molarity n or n Molarity ? Volume V
(2.5 ? 10-2 L) ? (0.75 M Ag)
1.9 ? 10-2 mol Ag
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Calculating mass from Molarity A chemist is
analyzing the alcohol content of a wine from
Napa. To do this she needs to use 1.00L of an
0.200 M K2Cr2O7 solution. (molar mass of K2Cr2O7
294.2 g mol-1) How much solid K2Cr2O7 must be
weighted out to make this solution?
Molarity of solution 0.200 M Volume of solution
1.00 L grams of K2Cr2O7 ?
Step 1 Calculate the moles of K2Cr2O7 present.
Molarity n or n Molarity ? Volume V
(1.00 L) ? (0.200 M K2Cr2O7) 0.200 mol
K2Cr2O7
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Step 2 Convert moles into grams
n mass or mass Molar mass ? n molar
mass
(0.200 mol K2Cr2O7) ? (294.2g K2Cr3O7)
58.8g K2Cr2O7
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Making a Standard Solution
A standard solution is a solution whose
concentration is accurately known.
This done by weighing out a sample of solute,
transferring it completely to a volumetric flask,
and adding enough solvent to bring the volume up
to the mark on the neck of the flask.
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Formulae for solution problems
n number of moles (mol) m mass (g) MM molar
mass (g mol-1)
M Molarity (mol L-1) n number of moles
(mol) V volume (L)
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Dilution
To save space in out Prep room we buy solutions
in concentrated form, i.e. 18M HCl (18 mol L-1).
We call these stock solutions.
The process of adding more solvent to a solution
is called dilution. A typical dilution involves
determining how much water must be added to an
amount of stock solution to achieve a solution of
the desired concentration.
The amount of solute after dilution moles of
solute before dilution
Remains constant
M moles of solute volume (L)
Decreases
Increases (water added)
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Example We want to prepare 500. mL of 1.00 M
acetic acid, CH3COOH, from a 17.5 M stock
solution of acetic acid. What volume of the stock
solution is required?
(b) water is added to dilute it
(c) final diluted solution has a volume of 0.5L
and a molarity of 1 M acetic acid
(a) What volume of 17.5 M acetic acid do you have
to remove?
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Example We want to prepare 500. mL of 1.00 M
acetic acid, CH3COOH, from a 17.5 M stock
solution of acetic acid. What volume of the stock
solution is required?
Step1 Find the number of moles of acetic acid
needed in final solution
n molarity of dilute solution ? volume of
dilute solution n 1.00 molL-1 ? 0.5 L n
0.500 mol CH3COOH
Step2 Find the volume of 17.5 M acetic acid that
contain 0.500 mol of CH3COOH. We will call this
unknown volume V.
V moles of solute 0.500 mol CH3COOH
Molarity 17.5 mol L-1
V 0.0286 L or 28.6 mL
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(a) 28.6 mL of 17.5 M acetic acid solution is
transferred to a volumetric flask that already
some water.
(b) water is added to the flask up to the 500 mL
mark.
(c) The final solution is 1.00 M acetic acid
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Because the moles of solute remain the same
before and after dilution, we can write
Final Conditions
Initial Conditions
M1 ? V1 moles of solute M2 ? V2
Volume before dilution
Volume after dilution
Molarity before dilution
Molarity after dilution
17.5 M ? 0.0286 L moles of solute 1.0 M ? 0.5
L
17.5 M ? 0.0286 L moles of solute 0.500
mol
1.0 M ? 0.5 L moles of solute 0.500 mol
M1 ? V1 moles of solute M2 ? V2
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Use this equation to solve this problem M1 ?
V1 M2 ? V2
What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of a 0.10 M H2SO4 solution?
M1 16 molL-1 M2 0.10 molL-1 V1 ? V2
1.5 L
9.4 ? 10-3 L 9.4 mL