Transcript Solution Concentrations

Solutions - Quantitatively
Solutions
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Mixture of at least two components
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Components can be any phase
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Solute
Solvent
Usually liquid solvent, but can be solid or gas
Examples ?
Quantifying Solution Components
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Concentration
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Ratio of solute to solvent present
Qualitative Descriptors: Concentrated, Dilute,
Saturated, Unsaturated
Ways to measure
Concentration
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Molarity
Molality
% by mass
Mole Fraction
Molarity
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Symbol = (M) or (M)
M = moles of solute
Liter of solution
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How to make a 1.0 M solution of NaCl:
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- Measure out 1.0 mole of NaCl, pour into a volumetric flask
- Add enough water to make exactly 1.0 L of solution
Practice Calculation
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Calculate the molarity of a solution that is made of 1.56
grams of gaseous HCl dissolved in enough water to
make 26.8 mL of solution.
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Step 1:
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Step 2:
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Step 3:
Calculate the number of moles of HCl
Convert the volume to Liters
Divide the moles of solution by the Liters of solution
Answer = 1.60 M HCl
More Practice…
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What is the molarity of a solution prepared by dissolving
11.5 g of NaOH in enough water to make 1.50 L sol’n?
Answer = 0.192 M
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If I want to make 1.00 L of a 0.200 M solution of
K2Cr2O7, how much solid potassium dichromate do I
need?
Answer = 58.8 g K2Cr2O7
How many moles of CuCl2 are in 250 mL of 3.0 M sol’n?
Answer = 0.75 moles
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What are the concentrations of all ions present in a 0.50
molar solution of Co(NO3)2? Answer = [Co+2] = 0.50 M
[NO3-1] = 1.0 M
Dilution Calculations
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Start with Concentrated Stock Solution of
known concentration
Add water to make diluted solution of
specific desired concentration
Example: Prepare 500. mL of 1.0 M HCl
from 12.0 M stock.
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Step 1:
Step 2:
Step 3:
Determine moles of HCl needed in final solution.
Determine volume of stock with that number of moles HCl.
Add enough water to stock to make volume needed.
Example: Prepare 500. mL of 1.0 M HCl
from 12.0 M stock.
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Step 1: Determine moles of HCl needed in final solution.
500. mL solution x
1L
1000 mL
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x 1.0 mole HCl = 0.500 moles HCl
1.0 L sol’n
Step 2: Determine volume of stock with that number of moles HCl.
12.0 mol HCl = 0.500 mol HCl
1.00 L sol’n
x L solution
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X = 0.042 L or 42 mL stock
Step 3: Add enough water to stock to make volume needed.
Add 458 mL water to the 42 mL of stock to make exactly 500. mL of 1.0 M HCl.
Moles of HCl in stock = Moles of HCl in dilution
Stoichiometry of Precipitation Rxns
Calculate the mass of solid NaCl that must be added to
1.50 L of a 0.100 M AgNO3 solution to precipitate all of the
Ag+ ions in the form of AgCl.
1: Write the balanced net ionic equation.
2: Determine the number of moles of Ag+ present.
3: Determine how many moles Cl-1 ions are needed to react all Ag+,
and therefore, how many grams NaCl that is.
Answer = 8.77 grams NaCl
Another example…
When sodium sulfate and lead (II) nitrate are mixed, a
precipitate forms. Calculate the mass of the precipitate
when 1.25 L of 0.0500 M lead (II) nitrate are mixed with
2.00 L of 0.0250 M sodium sulfate.
1: Write the balanced equation, then the net ionic equation.
2: Calculate the moles of each reactant that produces precipitate.
Determine which is limiting.
3. Calculate the number of moles of product formed. Convert to grams.
Answer = 15.2 grams of PbSO4
Acid/Base Reactions
What volume of a 0.100 M HCl solution is needed to
neutralize 25.0 mL of 0.350 M NaOH solution?
1: Write the balanced equation, then the net ionic equation.
2: Calculate the moles of each reactant.
Determine which is limiting.
3: Calculate moles of reactant needed, then convert to Volume.
Answer = 8.75 x 10-2 L or 87.5 mL 0.100 M HCl
Acid/Base Titrations
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Quantitative Acid/Base Neutralizations
Used to determine unknown concentration of acid or base.
Indicator is added so that endpoint/equivalence point of
reaction is seen.
Acid/Base Titrations
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For calculation of the unknown concentration to be
successful, you must :
 Know the exact reaction happening.
 Be able to accurately reach the equivalence point
experimentally.
 Be able to determine the exact volume of titrant
added to reach the equivalence point.
M(H+)V(H+) = M(OH-)V(OH-)
Equivalence Point
Phenolphthalein Indicator
Before Eq. Pt.
At Eq. Pt
Overtitrated
Graphical Analysis of Titration
Titration Calculation
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M(H+)V(H+) = M(OH-)V(OH-)
A 25.00 mL sample of HCl (aq) requires 24.16 mL of
0.106 M NaOH for complete neutralization. What is the
concentration of the original hydrochloric acid solution?
Answer = 0.102 M HCl