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# Nyb F09 Unit 1 Slides 37 73

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• Reaction of calcium chloride with sodium carbonate.
• Quantitative transfer typically means to measure a solid in another vessel than its final destination, and transferring the solid in its entirety to the final vessel. This is usually done by dissolving the solid with a solvent (usually water), transferring, and then rinsing the vessel and transferring repeatedly to the final vessel. Common measuring vessel: a beaker (wide mouth makes for easy access) Common final vessel: volumetric flask (very narrow opening makes adding solids into it directly difficult)
• A 1:6 dilution is made of a 2.25 M solution. What is the concentration of the dilution? Explain how you would make a 1 in 20 dilution of a 1.50 M solution using a 500 ml volumetric flask. What is the concentration of the dilution? A diluted solution is 0.750 M, which was made by diluted a stock solution 15X. What is the concentration of the original stock solution?
• A stock solution is 1.25 M. If it has been diluted 6X, what is the concentration of the dilution? 1.25 M/6 = 0.208 M You have unlimited 2.5 M stock solution and are asked to make 1.0 L that is 20X more dilute. How would you do this? The 2.5 M that is 20X less concentrated is 0.125 M. The quick way to calculate this is to realize that the volume required is 20 times less than the desired volume. 1.0 L/ 20 = 0.050 L Take 50 ml of stock and place in 1 L volumetric flask. Add water to the line and mix.
• Mg(s) + O2(g)  MgO(s) C8H18(l) + O2(g)  CO2(g) + H2O(g) NCl3(g) + H2O(l)  NH3(g) + HOCl(aq)
• Give examples of dissociation stoichiometry: NaCl(aq)  Na+(aq) + Cl-(aq)
• Making Ic + Ch 8 20 2Ic + Ch 16 10 8Ic + 5Ch 40 30
• NaCl(aq)  Na+(aq) + Cl-(aq) 1 mole ? ? Or IRF method CaCl2(aq)  ions?
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 1 mole methane burning in excess air (oxygen) 38.0 g collected. What is percent yield. In an experiment, 35.00 ml of 0.20 M silver nitrate(aq) was mixed with 45.00 ml of 0.25 M potassium bromide(aq) to produce a precipitate. a) Write a balanced formula for the reaction b) Calculate the theoretical yield (in grams) of the precipitate that was produced in the reaction c) Calculate the molarity of each of the ions present in the final solution
• ON of zinc chloride, sulfur trioxide, nitric acid 2Al + 3H2SO4  Al2(SO4)3 + H2(g) PbO + CO  Pb(s) + CO2(g)
• 2Li(s) + 2H2O(l)  2LiOH(aq) + H2(g) Ni(s) + HCl(aq)  NiCl2(aq) + H2(g) Zn(s) + CuSO4(aq)  Cu(s) + ZnSO4(aq) Cl2(aq) + 2NaBr(aq)  2NaCl(aq) + Br2(aq) Predicting Reactions
• Predicting reactions Reaction will not occur if reaction on top is to be reversed. Reactions proceed when top reaction moves forward, and bottom reaction becomes reversed
• ZnlZn2+ ll Cu2+l Cu Oxidation on left, reduction reaction right Phases separated by single vertical line, double vertical separates the 2 half cell reactions represents salt bridge When would this set up stop producing electricity? If light bulb removed, and AC power supply connected to middle, it will drive non-spontaneous reverse reaction (as long as greater than 1.10 V). Same concept of how rechargeable batteries work.
• Notice typically no salt bridges on electrolytic cells. Both electrodes are in the same solution.
• Notice no salt bridge. Salt must be molten for ions to move, allowing current to flow. For NaCl it’s 801 oC Spontaneous reaction: 2 Na(s) + Cl2(g)  2 NaCl(s) (split into half reactions) Non-spontaneous: NaCl(s)  Na(s) + Cl2(g) (split into half reactions at each electrode) Why not do this in aqueous NaCl
• The Na+ cannot take the electrons since it takes more energy than for H2O. The Cl- will not lose the electrons since it too take more energy than for H2O to lose electrons.
• 1. CH3CH2OH + O2  CO2 + H2O 2. 2 H2O2  2 H2O + O2
• ### Transcript

• 1. • Precipitation Reaction (Double Displacement): Solid precipitate is formed from mixing two aqueous solutions. Clear aqueous Solution #1 Clear aqueous Solution #2 Mix The cloudy, powdery solid formed is called a precipitate.
• 2. 1. Molecular Equation ( Formula Equation): Written as if everything still existed as compounds in aqueous solution. We know that soluble ionic compounds form ions, so this equation is often the least useful , but it is written in this manner out of convenience. 2. Complete (Total) Ionic Equation: Written such that all chemical species exist in their “true” state in solution. This equation is the most representative of what is really happening in solution, but it is the longest and most tedious to write out. 3. Net Ionic Equation: This equation represents the chemistry that occurred in solution , and is basically the total ionic equation without the spectator ions. • Representing Reactions: (Mainly in Aqueous Solutions)
• 3. Solution Concentration (Ch. 4) Mol ar ity (M) Mol al ity ( m )
• Standard, most common
• measurement of solution
• concentration
- Can be affected by temperature changes
• Far less commonly used
• compared to M
• Used when temperature
• changes may be an issue
Eg. A 1.0 molal solution 1.0 m 1.0 mol/kg Eg. A 1.0 molar solution 1.0 M 1.0 mol/L Moles of solute liter of solvent Moles of solute kilogram of solvent
• 4. • Making Molar Solutions
• Molar solutions are made of certain amount of moles of solute
• in an absolute volume of solvent (water)
20.000 g of NaCl (has “volume”) 250 mL Volumetric Flask - Place solid in volumetric flask of desired total volume (or use quantitative transfer ) - Add water to the line (exactly 250.00 mL total volume), cap, and mix to dissolve completely
• Proper glassware is required for preparing molar solutions
• 5.
• Molal solutions are easier to make than molar solutions
- An accurate balance is essential since all measurements are based on mass Exact mass of solute (say it’s 20.00 g of NaCl) Exact mass of solvent (say it’s 250.00 g of water)
• After mixing,
• exact molal
• concentration
• can be easily
• calculated
• Making Molal Solutions
• 6. - To make a dilution from an original “stock” solution, you need to add water - The amount of stock you use , and the total volume of the new diluted solution need to be known accurately in order to make accurate dilutions 1. Say you have a stock solution that is 1.0 M. You take 50 ml of the solution, and add exactly 50 ml of pure water to it. What is the concentration of the dilution? Example: Intuitively, you probably realize that by doubling the volume, the diluted solution is now half as concentrated  0.50 M To do calculations, we use a simple formula: • Making Dilutions
• 7. • “ Formula” to calculate dilutions: M 1 V 1 = M 2 V 2 Concentration of original stock solution (mol/L) Volume of stock solution that we are diluting (L) New total volume of dilution (L) New concentration of dilution (mol/L) (sometimes the formula is seen as C 1 V 1 = C 2 V 2 ) Time Saver: You may leave the V’s in units of mL (moles before) (moles after)
• 8. Solution/Dilution Problems: a) 15.00 ml of 1.00 M stock solution is placed in a 50 ml volumetric flask and filled up to the line with dH 2 O and mixed. What is the concentration of this new solution? b) You want to make 500 ml of a 0.025 M solution of NaCl, but there is no more solid salt. You will have to use the 1.00 M stock solution. Explain how you would make the solution. c) 25.00 ml of a 1.00 M solution, and 75.00 ml of a 0.0250 M solution are mixed together. What is the concentration of this new solution?
• 9. - There are various terms for dilutions: Eg. Dilute the solution by 10 fold Make a 10 X dilution Make a 1 in 10 dilution The solution was diluted 1:10 - For example, here are various terms you may see describing a diluted solution that is 10 times less concentrated that the original In this particular example, the dilution factor is 10 • Dilution Factors - In a “practical sense” “1 in 10” dilution means to take 1 part of stock and mixing accurately with 9 parts water .
• 10. 2.) Explain how you would make a 1 in 20 dilution of a 1.50 M solution using a 500 ml volumetric flask. What is the concentration of the dilution? 3.) A diluted solution is 0.750 M, which was made by diluted a stock solution 15X. What is the concentration of the original stock solution? 1.) A 1:6 dilution is made of a 2.25 M solution. What is the concentration of the dilution?
• 11. Balancing Equations and Stoichiometry (Ch. 4) • Chemical equations must have the same number of atoms on either side of the reaction equation, leading to a balanced equation . In chemistry, having the proper ratio of atoms is referred to as stoichiometry , meaning a measure of an element or part. • Generally, we always want to balance an equation using the smallest whole-number coefficients, but sometimes fractional coefficients do come in handy. 1. If an element occurs in only one compound on each side of the equation, balance this element first . Two Quick Tips For Balancing Equations: 2. When one of the reactants or products exists as the free element , balance this element last .
• 12. • Stoichiometry with Actual Quantities of Chemicals - The stoichiometry of a balanced equation dictates the proper ratios of reactants to products. Eg. Ca 2+ (aq) + 2 OH - (aq)  Ca(OH) 2 (s) - In this example, 1 ion of calcium must react with 2 ions of hydroxide to form one ionic unit of calcium hydroxide Questions: a) How many OH - ions must 1 million calcium ions react with to form 1 million units of calcium hydroxide? b) The reaction above just formed exactly 2 moles of calcium hydroxide. How many moles of the two ions must have reacted together? CHEMICALS REACT ON A MOLE-TO-MOLE BASIS
• 13. Stoichiometry with Food: Ice Cream Sundaes Ic Ch Scoop of ice cream Cherry The periodic table of food: • One can think of chemical equations as a recipe to form products where the ratio of reactants to products is fixed!
• 14. The IRF Table to Keep Track of Reaction Stoichiometry • I stands for Initial . In this line, the initial quantities of substances in moles are entered • R stands for Reaction . In this line, the reactants are used up so the quantities are represented with a minus sign . The products are being formed so the quantities are represented with a plus sign . • F stands for Final . This line represents the final quantities of all substances. The F line is the I line + the R line . The limiting reagent (reactant): This is the reactant that is completely used up and limits how much product can be formed. It always goes to 0 in the F line .
• 15. A Bookkeeping Tip For IRF Tables:
• Determine the I nitial molar amounts of your reactants and write them into line I.
2. If the stoichiometry of the reactants is not 1-to-1, leave a space between line I and line R, and call it the “ Standarization line ”, or (s). Divide the molar quantity of each reactant by its corresponding stoichiometric coefficient and write these new numbers in the (s) line.
• Compare the numbers in the (s) line. The smallest number is
• the limiting reagent. Box, circle, or mark this number. Ignore
• or cross off the other (s) line numbers.
• Use the S line to determine the R line. This is done by
• mulitiplying the limiting reagent number (boxed) in the (s) line
• by the corresponding stoichiometric coefficient for each
• reactant or product and writing the number into the R line.
• Continue as usual to determine the F line
• 16. • The quantity of product formed in the F line is called the theoretical yield . This is how much of the product we expect to get if the reaction was 100% efficient. • Experimentally, reactions are never 100% efficient and product is lost somewhere along the way (by-products of reaction, lost during isolation, etc.). The actual amount of product recovered is called the actual yield . Actual Yield Theoretical Yield X 100 = Percent Yield
• 17. 1.) In an experiment, 35.00 ml of 0.200 M silver nitrate solution was mixed with 45.00 ml of 0.250 M potassium bromide solution to produce a precipitate. a) Write a balanced equation for the reaction b) Calculate the theoretical yield (in grams) of the precipitate that was produced in the reaction c) Calculate the molarity of each of the ions present in the final solution • Stoichiometry of Precipitation Reactions
• 18. 2.) 12.47 g of solid sodium phosphate has been dissolved in 150.00 ml of water. In an experiment, 100.00 ml of 0.650 M magnesium nitrate solution was mixed with the sodium phosphate solution to form a precipitate. a) Calculate using an IRF table the theoretical yield (in grams) of the precipitate that could be produced from the reaction b) If 4.61 g of ppt was recovered, what is the percent yield? c) Calculate the molarity of all remaining ions present in the final solution
• 19. • Acid-Base Reaction: A proton (H + ) is transferred from one chemical species to another. Often, an acid and base in solution react to produce water and a salt, a process called neutralization . HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) Eg. HCl(aq) + NH 3 (aq)  NH 4 + (aq) + Cl - (aq) • Classical (Arrhenius) Definitions of Acids and Bases - Acid: Substance that produces H + ions in aqueous solution - Base: Substance that produces OH - ions in aqueous solution Eg. HCl, CH 3 COOH, H 2 SO 4 Eg. NaOH, KOH
• 20. • Titration: A substance in solution of known concentration is slowly reacted with another substance in solution of unknown concentration. - If one knows the stoichiometry of the reaction occuring, one can determine the concentration of the solution with previously unknown concentration - In an acid-base titration the typical net ionic reaction occuring is: H + (aq) + OH - (aq)  H 2 O(l) - Care must be taken with acids that have more than 1 “ acidic” proton, and bases that can generate more than 1 equivalent of OH - .
• 21. • Titration Technique: The standard lab equipment used for titration is the burette . A burette allows for slow addition of titrant (solution of known concentration) into the analyte solution (unknown concentration) - Equivalence Point: When the moles of acid equals the molar equivalent of base OR when the moles of H + equals the moles of OH - . (A chemical phenomenon; cannot see it) - End Point: When an indicator changes colour, signifying the end of the titration. (A physical phenomenon; must be observed) - Acid-Base Indicators: A dye whose colour changes depending on the acidity or basicity of the solution
• 22. Acid-Base Indicators
• 23.
• How many moles of sodium hydroxide would it take to
• fully neutralize 70.0 ml of 0.15 M HCl?
2. You have titrated 25.0 ml of HCl of unknown concentration. It took 46.24 ml of 0.24 M NaOH to reach the end point. What is the concentration of the HCl solution? 3. How much 0.16 M sulfuric acid would it take to fully neutralize 35.0 ml 0.55 M potassium hydroxide? 4. You have titrated 10.0 ml of sulfuric acid solution of unknown concentration. It took 33.23 ml of 0.45 M NaOH to reach the end point. What is the concentration of the sulfuric acid solution?
• 24.
• Oxidation is the loss of electrons
• - Reduction is the gain of electrons.
- An oxidizing agent , is a chemical species that is doing the oxidizing, so that it itself is reduced . - A reducing agent is a chemical species that is doing the reducing, so that it itself is oxidized . • Oxidation-Reduction (Redox) Reaction: Electrons are transferred from one reactant to another. OIL RIG
• 25. • Oxidation Numbers (Simple): O.N. (oxidation number) - An element 0 - An individual ion same as its charge As part of a compound: - Group 1A and hydrogen atoms +1 - Group 2A atoms +2 - Oxygen atoms -2 - Halogen atoms (group 7A) -1 Eg. 2 Na(s) + Cl 2 (g)  2 NaCl(s) 0 0 +1 -1 Net charge must be equal on both sides of the equation
• 26. - Single Displacement Reaction: a type of redox reaction where a single substance (usually a metal or halogen) reacts with a compound in solution to form a gas or new element . a.) metal in water or acidic solution produces H 2 gas b.) metal in solution produces a new metal c.) halogen in solution produces a new halogen Zinc metal placed in copper sulfate solution
• 27. In the previous example: Oxidation: Zn(s)  Zn 2+ (aq) + 2 e - Reduction: Cu 2+ (aq) + 2 e -  Cu(s)
• 28. Electrochemistry (Ch. 18) • Electrochemistry is the study of chemical reactions occuring in solution that allow for electron movement between electrodes (electric conductors) Involves Redox Chemistry • One of the greatest benefit of electrochemistry is the creation of batteries and fuel cells. Through feats of engineering, a redox chemical process occurs in a self contained device that creates electricity upon command. • In electrolysis, compounds are broken back down into their elemental forms. Electro = electricity Lysis = splitting
• 29. • A Voltaic Cell: Generates electricity from spontaneous chemical reactions Cell Potential or E cell = E o cathode - E o anode Voltage = 1.10 V
• 30.
• 31. • An Electrolytic Cell: Uses energy to drive non-spontaneous chemical reactions  electrolysis Molten NaCl(l) Power Supply - + Na(s) Cl 2 (g) Over time Electron flow Molten NaCl(l) Power Supply - + Molten Salt Electrolytic Cell
• 32. Electrolysis of Water Ions of a salt such as NaCl must be added so current can be carried through the water. - 0.40 V - 0.83 V Na + and Cl - ions also present in solution
• 33. • Extra Reactions: 1. Combustion Reaction: a substance combines with oxygen to produce one or more oxygen-containing compounds 3. Acid and Carbonate: when a source of H + reacts with a carbonate species ( CO 3 2- or HCO 3 - ), carbon dioxide and water are formed. - the release of carbon dioxide is so rapid that the reaction bubbles vigorously 2. Decomposition Reaction: a single compound breaks down into elements or simpler compounds.
• 34. Solutions and Electrical Conductivity • Electrical conductivity is the measure of the ability of a material to carry electrical current • Solutions can conduct electricity as long as they contain large quantities of electrolytes (ions in solution) • The ions in solution “polarize” by moving to the proper electrode, allowing for electrical current to flow
• 35.
• 36.
• Soluble ionic compounds will dissolve in water and
• dissociate completely (100%) into its constituent ions
• forming a high concentration of ions in solution, and are
• therefore strong electrolytes .
- A slightly soluble ionic compound would generate small amounts of ions in solution and are therefore weak electrolytes .
• Insoluble ionic compounds would not dissolve and
• there would be no ions in solution and would be
• considered a non-electrolyte .
Electrolytes • Ionic Compounds
• 37. - Most molecular compounds do not dissociate when dissolved in water due to the strength of their covalent bonds. - Since no ions are formed, most molecular compounds are considered non-electrolytes . Exceptions: a) Strong acids are molecular compounds that dissolve completely into ions and are considered strong electrolytes . (There are 6 strong acids) b) Weak acids and weak bases are molecular compounds that form small amounts of ions in solution and are considered weak electrolytes . • Molecular (Covalent) Compounds