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Chemical Reactions
in Solutions
Dr. K. Shahzad Baig
Memorial University of Newfoundland
(MUN)
Canada
Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario.
Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.
Chemical Reactions in Solution
Solution: homogeneous mixture of two or more substances.
Solute: the substance being dissolved.
Solvent: the substance doing the dissolving.
Generally, the component present in the greatest
quantity is considered to be the solvent.
Aqueous solutions are those in which water is the solvent.
Copyright McGraw-Hill 2009 3
Comparison of a Concentrated and Dilute Solution
Molarity
π‘šπ‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ =
π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ (𝑖𝑛 π‘šπ‘œπ‘™π‘’π‘ )
π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› (𝑖𝑛 π‘™π‘–π‘‘π‘’π‘Ÿπ‘ )
The composition of a solution may be specified by giving its molarity
𝑀, (π‘šπ‘œπ‘™/𝐿) =
𝑛 (π‘šπ‘œπ‘™)
𝑉 (π‘™π‘–π‘‘π‘’π‘Ÿπ‘ )
If 0.440 mol urea, is dissolved in enough water to make 1.000 L of solution, the solution
concentration, or molarity
=
0.440 π‘šπ‘œπ‘™ 𝐢𝑂 𝑁𝐻2 2
1.000 𝐿 π‘ π‘œπ‘™π‘›
= 0.440 𝑀𝐢𝑂 𝑁𝐻2 2
Problem statement
How many grams of NaCl is required to make 0.5 L of 0.5 molar NaCl
58.44 π‘”π‘Ÿπ‘Žπ‘šπ‘  π‘π‘ŽπΆπ‘™
1 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
π‘₯ 0.5 𝐿 π‘₯ 0.5
π‘šπ‘œπ‘™
𝐿
= 14.61 π‘”π‘Ÿπ‘Žπ‘šπ‘ 
A solution is prepared by dissolving 25.0 mL ethanol, in enough
water to produce 250.0 mL solution. What is the molarity of
ethanol in the solution?
Example 4-8
A solution is prepared by dissolving 25.0 mL ethanol, CH3CH2OH (d = 0.789 g/mL), in
enough water to produce 250.0 mL solution. What is the molarity of ethanol in the solution?
= 1.71 M CH3CH2OH
π‘šπ‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ =
π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ (𝑖𝑛 π‘šπ‘œπ‘™π‘’π‘ )
π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› (𝑖𝑛 π‘™π‘–π‘‘π‘’π‘Ÿπ‘ )
Problem statement
What is the molality of acetone in a solution composed of 255 g of acetone, (CH3)2CO,
dissolved in 200 g of water? The molar mass of acetone is 58.08β‹…gβ‹…molβˆ’1,
π‘€π‘œπ‘™π‘Žπ‘™π‘–π‘‘π‘¦ =
π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
𝐾𝑔 π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘
π‘€π‘œπ‘™π‘Žπ‘™π‘–π‘‘π‘¦ = 225 𝑔 π‘₯
1 π‘šπ‘œπ‘™
58.08 𝑔 π΄π‘π‘’π‘‘π‘œπ‘›π‘’
π‘₯
1
200 𝑔
π‘₯
1000 𝑔
1 π‘˜π‘”
= ? ? mol / kg
Problem statement
A 22.3 g sample of acetone (see the model here) is dissolved in enough water to
produce 1.25 L of solution. What is the molarity of acetone in this solution?
Molarity (M) = the number of moles of solute per liter of solution
Molality (m) = the number of moles of solute per kilogram of solvent.
π‘€π‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ =
π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
𝐿 π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
π‘€π‘œπ‘™π‘Žπ‘™π‘–π‘‘π‘¦ =
π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
𝐾𝑔 π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘
Example 4-9
What mass of is needed to prepare exactly 0.2500 L (250.0 mL) of a 0.250 M solution in H2O?
L soln >> mol K2CrO4 >> g K2CrO4The conversion pathway is
Solution
Dilution
𝑀 (π‘šπ‘œπ‘™/𝐿) =
𝑛 (π‘šπ‘œπ‘™)
𝑉 (π‘™π‘–π‘‘π‘’π‘Ÿπ‘ )
𝑛 π‘šπ‘œπ‘™ = 𝑀 π‘€π‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ π‘₯ 𝑉 (πΏπ‘–π‘‘π‘’π‘Ÿπ‘ )
𝑛1 = 𝑀1 π‘₯ 𝑉1 𝑛 𝑓 = 𝑀𝑓 π‘₯ 𝑉𝑓
𝑀1 π‘₯ 𝑉1 = 𝑀𝑓 π‘₯ 𝑉𝑓
𝑛1 = 𝑛 𝑓
Problem Statement
A particular analytical chemistry procedure requires 0.0100 M K2CrO4.
What volume of 0.250 M must be diluted with water to prepare 0.2500 L of 0.0100
K2CrO4 ?
𝑀1 π‘₯ 𝑉1 = 𝑀2 π‘₯ 𝑉2
𝑉1 = 𝑉2 π‘₯
𝑀2
𝑀1
𝑉1 = 250.0 π‘šπΏ π‘₯
0.0100 𝑀
0.250 𝑀
= 10.0 mL of 0.250 M K2CrO4
Problem statement
A 25.00 mL pipetful of 0.250 M K2CrO4 is added to an excess of AgNO3 (aq).
What mass of will precipitate from the solution?
K2CrO4 (aq) + 2 AgNO3 (aq) β†’ Ag2CrO4 (s) + 2 KNO3 (aq)
First convert the volume of K2CrO4(aq) from milliliters to liters, and then
use molarity as a conversion factor between volume of solution and moles of solute
Use the molar mass to convert from moles to grams of Ag2CrO4 .
Use a stoichiometric factor from the equation to convert from moles of K2CrO4 to
moles of Ag2CrO4 .
Determining the Limiting Reactant
When all the reactants are completely and simultaneously consumed in a chemical
reaction, the reactants are said to be in stoichiometric proportions
A limiting reactant is the substance
that is totally consumed when the
chemical reaction is complete.
The amount of product formed is
limited due to this reactant, since the
reaction cannot continue without it.
EXAMPLE 4-12
Phosphorus trichloride, is a commercially important compound used in the manufacture of
pesticides, gasoline additives, and a number of other products. Liquid is made by the direct
combination of phosphorus and chlorine.
P4 (s) + 6 Cl2 (g) β†’ 4 PCl3(l)
What is the maximum mass of PCl3 that can be obtained from 125 g P4 and 323 g Cl2 ?
Solution
Step 1. Find moles of reactants ,
Step 2. Compare the mole ratio
Since there is less than 6 mol Cl2 per mole of P4, chlorine is the limiting reactant
P4 (s) + 6 Cl2 (g) β†’ 4 PCl3(l)
Convert from moles of PCl3 to grams of PCl3 by using the molar mass of PCl3.
Convert from moles of Cl2 to moles of PCl3 by using the stoichiometric factor
Problem statement
What would be the limiting reagent if 26.0 grams of C3H9N were reacted with
46.3 grams of O2?
4C3H9N+25O2 β†’ 12CO2+18H2O+4NO2
From C₃H₉N: considering CO2 is a product
From Oβ‚‚:
The Oβ‚‚ gives the smaller amount of COβ‚‚, so the Oβ‚‚ is the limiting reactant.

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Chemical reactions in solutions

  • 1. Chemical Reactions in Solutions Dr. K. Shahzad Baig Memorial University of Newfoundland (MUN) Canada Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario. Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.
  • 2. Chemical Reactions in Solution Solution: homogeneous mixture of two or more substances. Solute: the substance being dissolved. Solvent: the substance doing the dissolving. Generally, the component present in the greatest quantity is considered to be the solvent. Aqueous solutions are those in which water is the solvent.
  • 3. Copyright McGraw-Hill 2009 3 Comparison of a Concentrated and Dilute Solution
  • 4. Molarity π‘šπ‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ = π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ (𝑖𝑛 π‘šπ‘œπ‘™π‘’π‘ ) π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› (𝑖𝑛 π‘™π‘–π‘‘π‘’π‘Ÿπ‘ ) The composition of a solution may be specified by giving its molarity 𝑀, (π‘šπ‘œπ‘™/𝐿) = 𝑛 (π‘šπ‘œπ‘™) 𝑉 (π‘™π‘–π‘‘π‘’π‘Ÿπ‘ ) If 0.440 mol urea, is dissolved in enough water to make 1.000 L of solution, the solution concentration, or molarity = 0.440 π‘šπ‘œπ‘™ 𝐢𝑂 𝑁𝐻2 2 1.000 𝐿 π‘ π‘œπ‘™π‘› = 0.440 𝑀𝐢𝑂 𝑁𝐻2 2
  • 5. Problem statement How many grams of NaCl is required to make 0.5 L of 0.5 molar NaCl 58.44 π‘”π‘Ÿπ‘Žπ‘šπ‘  π‘π‘ŽπΆπ‘™ 1 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™ π‘₯ 0.5 𝐿 π‘₯ 0.5 π‘šπ‘œπ‘™ 𝐿 = 14.61 π‘”π‘Ÿπ‘Žπ‘šπ‘  A solution is prepared by dissolving 25.0 mL ethanol, in enough water to produce 250.0 mL solution. What is the molarity of ethanol in the solution?
  • 6. Example 4-8 A solution is prepared by dissolving 25.0 mL ethanol, CH3CH2OH (d = 0.789 g/mL), in enough water to produce 250.0 mL solution. What is the molarity of ethanol in the solution? = 1.71 M CH3CH2OH π‘šπ‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ = π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ (𝑖𝑛 π‘šπ‘œπ‘™π‘’π‘ ) π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› (𝑖𝑛 π‘™π‘–π‘‘π‘’π‘Ÿπ‘ )
  • 7. Problem statement What is the molality of acetone in a solution composed of 255 g of acetone, (CH3)2CO, dissolved in 200 g of water? The molar mass of acetone is 58.08β‹…gβ‹…molβˆ’1, π‘€π‘œπ‘™π‘Žπ‘™π‘–π‘‘π‘¦ = π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ 𝐾𝑔 π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘ π‘€π‘œπ‘™π‘Žπ‘™π‘–π‘‘π‘¦ = 225 𝑔 π‘₯ 1 π‘šπ‘œπ‘™ 58.08 𝑔 π΄π‘π‘’π‘‘π‘œπ‘›π‘’ π‘₯ 1 200 𝑔 π‘₯ 1000 𝑔 1 π‘˜π‘” = ? ? mol / kg
  • 8. Problem statement A 22.3 g sample of acetone (see the model here) is dissolved in enough water to produce 1.25 L of solution. What is the molarity of acetone in this solution? Molarity (M) = the number of moles of solute per liter of solution Molality (m) = the number of moles of solute per kilogram of solvent. π‘€π‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ = π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ 𝐿 π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› π‘€π‘œπ‘™π‘Žπ‘™π‘–π‘‘π‘¦ = π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’ 𝐾𝑔 π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘
  • 9. Example 4-9 What mass of is needed to prepare exactly 0.2500 L (250.0 mL) of a 0.250 M solution in H2O? L soln >> mol K2CrO4 >> g K2CrO4The conversion pathway is
  • 10. Solution Dilution 𝑀 (π‘šπ‘œπ‘™/𝐿) = 𝑛 (π‘šπ‘œπ‘™) 𝑉 (π‘™π‘–π‘‘π‘’π‘Ÿπ‘ ) 𝑛 π‘šπ‘œπ‘™ = 𝑀 π‘€π‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ π‘₯ 𝑉 (πΏπ‘–π‘‘π‘’π‘Ÿπ‘ ) 𝑛1 = 𝑀1 π‘₯ 𝑉1 𝑛 𝑓 = 𝑀𝑓 π‘₯ 𝑉𝑓 𝑀1 π‘₯ 𝑉1 = 𝑀𝑓 π‘₯ 𝑉𝑓 𝑛1 = 𝑛 𝑓
  • 11. Problem Statement A particular analytical chemistry procedure requires 0.0100 M K2CrO4. What volume of 0.250 M must be diluted with water to prepare 0.2500 L of 0.0100 K2CrO4 ? 𝑀1 π‘₯ 𝑉1 = 𝑀2 π‘₯ 𝑉2 𝑉1 = 𝑉2 π‘₯ 𝑀2 𝑀1 𝑉1 = 250.0 π‘šπΏ π‘₯ 0.0100 𝑀 0.250 𝑀 = 10.0 mL of 0.250 M K2CrO4
  • 12. Problem statement A 25.00 mL pipetful of 0.250 M K2CrO4 is added to an excess of AgNO3 (aq). What mass of will precipitate from the solution? K2CrO4 (aq) + 2 AgNO3 (aq) β†’ Ag2CrO4 (s) + 2 KNO3 (aq) First convert the volume of K2CrO4(aq) from milliliters to liters, and then use molarity as a conversion factor between volume of solution and moles of solute
  • 13. Use the molar mass to convert from moles to grams of Ag2CrO4 . Use a stoichiometric factor from the equation to convert from moles of K2CrO4 to moles of Ag2CrO4 .
  • 14. Determining the Limiting Reactant When all the reactants are completely and simultaneously consumed in a chemical reaction, the reactants are said to be in stoichiometric proportions A limiting reactant is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited due to this reactant, since the reaction cannot continue without it.
  • 15. EXAMPLE 4-12 Phosphorus trichloride, is a commercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products. Liquid is made by the direct combination of phosphorus and chlorine. P4 (s) + 6 Cl2 (g) β†’ 4 PCl3(l) What is the maximum mass of PCl3 that can be obtained from 125 g P4 and 323 g Cl2 ? Solution Step 1. Find moles of reactants , Step 2. Compare the mole ratio
  • 16. Since there is less than 6 mol Cl2 per mole of P4, chlorine is the limiting reactant P4 (s) + 6 Cl2 (g) β†’ 4 PCl3(l)
  • 17. Convert from moles of PCl3 to grams of PCl3 by using the molar mass of PCl3. Convert from moles of Cl2 to moles of PCl3 by using the stoichiometric factor
  • 18. Problem statement What would be the limiting reagent if 26.0 grams of C3H9N were reacted with 46.3 grams of O2? 4C3H9N+25O2 β†’ 12CO2+18H2O+4NO2 From C₃H₉N: considering CO2 is a product From Oβ‚‚: The Oβ‚‚ gives the smaller amount of COβ‚‚, so the Oβ‚‚ is the limiting reactant.

Editor's Notes

  1. Concentrated – higher ratio of solute to solvent Dilute - smaller ratio of solute to solvent
  2. For molarity, L of solution is given. Convert given 22.3 grams of acetone to moles of acetone . Place the values in the Molarity foemula.
  3. that is, they are present in the mole ratios dictated by the coefficients in the balanced equation
  4. Ethylmethylamine