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Unit 9: Chemicals in (liquid) solutions -- general solutions, aqueous solutions --weak/strong electrolytes … acid/base/salt -- net ionic reactions Unit 10: Oxidation/Reduction Reactions -- exchange of electrons -- concentrations, Molarity Issues: ? Homework Set 3 due on Saturday @ 08:00AM EXAM 1 on MONDAY, Sept. 24 @ 7:15 PM People with a conflict: MUST sign up for alternate exam in Room 185 Chemistry by tomorrow
EXAM 1 on next MONDAY, Sept. 24 @ 7:15 – 8:15 PM You will receive an e’mail with your room assignment for the exam. Bring to the exam: 1) A photo-ID (msu id card, driver’s license, etc.) 2) Any stand-alone calculator provided that no chemical information is
stored in the calculator but NOT a cell phone, iPod, iPad, CPU, etc. 3) A pencil ! You will get: 1) An exam booklet (for you to keep). 2) With a periodic table & the constants/conversion factors/SI prefixes 3) Computer scoring sheet (we get to keep). 4) One hour, exam starts at 7:15pm, be in your seat then. 5) No one will be allowed to enter after the first person leaves.
(Aqueous) Acid/Base reactions often occur where some ions are unchanged by the process. These “spectator ions” are present before and after the chemical reaction and keep the solution electrically neutral. The “participant ions” carry out the reaction that can be described as the net reaction. e.g., KCl (s) + H2O (l) à K+ (aq) + Cl- (aq) [strong electrolyte] Possible synthesis: HCl (aq) + KOH (aq) à KCl (aq) + H2O (l) H+ + Cl- + K+ + OH- à K+ + Cl- + H2O (l) Net Reaction: H+ + OH- à H2O (l) Spectators: Cl- + K+ à K+ + Cl-
How would we know that ionic reactions were actually taking place if the compounds are dissociated and the ions are moving independently in the solution? Acid/Base Neutralizations, add NaOH to acid solution: HCl (aq) + NaOH (s) à NaCl (aq) + H20 (l) H+ + Cl- + NaOH (s) à Na+ + Cl- + H20 (l) H+ + OH- à H2O Net Reaction Monitor H+ concentration (pH .. More on this later in course)
How would we know that ionic reactions were actually taking place if the compounds are dissociated and the ions are moving independently in the solution? Precipitation Reactions, add NaCl solution to silver solution: AgNO3 (aq) + NaCl (aq) à AgCl (s) + NaNO3 (aq) Ag+ + NO3
- + Na+ + Cl- à AgCl (s) + Na+ + NO3-
Ag+ + Cl- à AgCl (s) Net Reaction Observe the formation of a solid from two colorless solutions
How would we know that ionic reactions were actually taking place if the compounds are dissociated and the ions are moving independently in the solution? Gas Formation Reactions, add NiCO3 (green powder) to acid solution: NiCO3 (s) + 2HCl (aq) à NiCl2 (aq) + H2O (l) + CO2 (g) NiCO3 (s) + 2H+ + 2Cl- à Ni2+ + 2Cl- + H2O + CO2 (g) NiCO3 + 2H+ à Ni2+ + H2O + CO2 (g) Net Reaction Observe the formation of bubbling gas leaving solution.
We need to keep track of the amount of material present in solutions for reactions, etc. Two options: mass per volume (density), or moles per volume Notice that “mass/volume” will need additional information to find the number of moles – let’s choose moles/volume !! Molarity (M) is simply the number of moles of a solute present in exactly one liter of the (mixed, total … ) solution. If I pour out some volume of a solution then the number of moles of solute I get is that volume times the molarity. Moles of solute = # of Liters * Molarity value of solution
Molarity (M) is simply the number of moles of a solute present in exactly one liter of the (mixed, total … ) solution. 1) Dissolve 49 g of H2SO4 in enough water to make 250 mL solution. What is M for
this mixture?
2) What mass of NaOH is required to make 15 L of 0.20 M solution?
3) How many mL of a 6.0M HCl solution need to be added to water to make 1.0L of a 0.15 M HCl solution?
Molarity = mass = 49 g = 2.0 M molar mass * volume 98.08 g/mol * 0.250L
Mass = g/mol * moles = g/mol * M * volume = 39.998 g/mol * 0.20 M * 15 L = 1.2x102 g
moles = M1 * volume1 = M2 * volume2 = 6.0 M * v = 0.15 M * 1.0 L à v = 0.15 M * 1. L / 6.0 M v = 0.025 L or 25 mL
In addition to the Acid/Base/Salt reactions there is another general class of chemical reactions that involves the transfer of electrons between atoms (or compounds) during the reaction. For historical reasons this group is called Oxidation-Reduction reactions and they can be quite dramatic, for example: 2 Mg (s) + CO2 (g) à 2 MgO (s) + C (s) element + oxide à metal oxide + element Oxidation: loss of electron(s) versus Reduction: gain of electron(s) We would like to have a scheme to keep track of the electron transfer.
We write oxidation numbers to indicate the distribution of electrons in chemical compounds and to be able to follow electron transfer in Oxidation-Reduction reactions. We will not count up all the electrons in an atom or molecule but use the difference from the neutral atom. The sum of oxidation numbers must equal the net charge on object. 1) An element in its native state is assigned to be zero. Na, Ne, O2 2) An atomic ion is assigned the value of the charge. La3+ is “+3” 3) Fluoride is always “-1” in compounds. 4) Group I elements are always “+1” in compounds.
Group II elements are always “+2” in compounds.
5) Oxygen always “-2” in compounds – except for OF2, O22-, O2
- 6) Cl, Br, and I always “-1” in compounds – except with F 7) Hydrogen always “+1” in compounds with nonmetals.
We write oxidation numbers to indicate the distribution of electrons in chemical compounds and to be able to follow electron transfer in Oxidation-Reduction reactions. We will not count up all the electrons in an atom or molecule but use the difference from the neutral atom. Examples from notes:
Mg3P2 Magnesium phosphide Mg +2, P -3 ClO2
- Chlorite anion -1 ion, O -2, Cl +3 SO4
2- Sulfate anion -2 ion, O -2, S +6
MnO4- Permanganate anion -1 ion, O -2, Mn +7
BrF3 Bromine trifluoride F -1, Br +3 CaH2 Calcium dihydride Ca +2, H -1 XeOF4 Xenon oxytetrafluoride F -1, O -2, Xe +6
Finally back to the example: 2 Mg (s) + CO2 (g) à 2 MgO (s) + C (s) element + oxide à metal oxide + element Oxidation numbers: Mg 0 Ox -2, C +4 à Ox -2, Mg +2 C 0 oxidation (increase) 2* [ Mg 0 à +2 ] reduction (decrease) C +4 à 0 Notice that the changes must balance (can’t create or destroy electrons)