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Exam 3 Study guide

by: Matthew Goetz

Exam 3 Study guide chem 10061-001

Matthew Goetz
GPA 3.925

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This has everything from chapter 17 until the buffer problems that were discussed.
general chemistry 2
David bowers
Study Guide
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This 5 page Study Guide was uploaded by Matthew Goetz on Tuesday April 19, 2016. The Study Guide belongs to chem 10061-001 at Kent State University taught by David bowers in Summer 2015. Since its upload, it has received 39 views. For similar materials see general chemistry 2 in Chemistry at Kent State University.


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Date Created: 04/19/16
Exam 3 Study  Chapter 17 focuses on the extent of a reaction.   To put this in simpler terms, we examined at what point in time a reaction reaches  equilibrium! Therefore, calculate at what point do they reach equilibrium (if they ever do  in the particular reaction we are examining), and how much product has formed once  equilibrium is reached.   At equilibrium, no net change will occur in the reaction’s concentrations.   Kc is equal to the concentrations of gaseous products divided by concentrations of  gaseous reactants. (This is the easiest way for me to express the equation that we use).   If K is very small then the reaction yields little product before reaching equilibrium. ­ This is described as the reaction favors reactants.  ­ If K is very large then it is the opposite.   Reaction Quotient (Q) ­ Ratio of products to reactants.  ­ This is calculated the same as K.  ­ However, they are different because Q may be calculated at any point in a  reaction and K may only be calculated at equilibrium!  ­ At equilibrium though, Q = K.  ­ If this isn’t true then the reaction must continue.  ­ If Q is smaller than K then the reaction must continue towards products. ­ If Q is larger than K then the reaction must continue towards reactants.   Once again, in these calculations we need not consider solids nor pure liquids because  their concentrations don’t change.   Kp is very similar to Kc, thought Kp deals with the partial pressures of the gases that are  present while Kc deals with the concentrations of them. They are calculated in the same  manner though.   A modified version of the ideal gas law allows us to change between Kc and Kp:  ­ Kp = Kc (RT)^Δn ­ R is a given constant, .0821 ­ T is the temperature in Kelvin ­ And n is the moles of product minus the moles of reactants.   If you are trying to solve for K or Q but you are missing more than one variable you must construct an ICE table!   An example of an ICE table is given here:   In this example, to solve for x you would make the top of the problem x^2, and then  bottom of the problem would be (.100­x) ^2). Since this is a perfect square you could  merely take the square root of the entire problem and then solve for x from there. Once  you have x you would plug it into the equilibrium part of the ICE table told solve for the  concentrations of each product and reactant at equilibrium.   However, if there is an X^2 and an X in your K equation, you must use the quadratic  formula to solve it.   The quadratic formula may be skipped though by assuming that the x that you subtract  from your initial concentration of reactants is so minimal that it will not be effective.  ­This assumption may only be made if your k is small and your initial concentration of  reactants is large.   If you do make this assumption, then you must also check it afterwards with the formula:  X/initial concentration times 100%.   If your answer is less than 5% then your assumption is good. If it is greater you must go and do the quadratic formula.   Le Chatelier’s Principle: If a change is induced to a system at equilibrium then the system will work to re­attain equilibrium by undoing the change.  ­For example, if more reactants are added then the system creates more products.  ­If more products are added then the system reduces them to reactants.  ­If the pressure is increased on a system then the side with the least amount of moles is  favored.  ­If the volume is increased then the pressure decreases and the side with more moles is  favored.  ­Adding temperature to the reactants causes more products to form.  ­Adding temperature to the products side of a reaction causes reactants to form.  ­Adding inert gasses to equilibrium don’t affect it, like solids or liquids.   An Arrhenius acid is an acid that dissociates an H+.  ­HCl + H2O  Cl­ + H3O+  Strong acids mostly dissociate, so very little reactants remain at equilibrium.   Same situation with bases as above stated.   The K constant for acids is Ka, and a high Ka means a stronger acid.   Water is treated as a constant when calculating Ka.  Kb is the constant for bases.      Strong Acids:  ­Hydrohalic acids ­Strong oxoacids, # of Oxygen is greater than or equal to # (O­H bonds) +2.   Weak Acids:  ­Weak oxoacids, # of Oxygen is lesser or equal to (O­H), or (O­H) +1.  ­HF is a weak acid.  ­All carboxylic acids are weak. ­Hydrated metal ions are weak.  ­Weak acids occur if H is not bound to an O or a halogen, like H2S, or HCN.   Strong bases:  ­Group 1 and 2 with –OH.   Weak Bases:  ­Ammonia, water, and any amines.   pH = ­log(H3O+)  pOH = ­log (OH­)  pKa = ­logKa  pKw = 14, or pH + pOH   pKa is the measure of the strength of acid dissociation.  Bronsted Lowry Acid:  ­Acid that donates H+. Same as Arrhenius.   Bronsted Lowry Base:  ­Base is an H+ acceptor. Means that a base must have a free electron pair.  ­Present in amines, ammonia, and fluorine.   From all of these acids and bases form conjugate acids and bases.   These are reaction specific situations, and an example is provided above.   The net direction of acid basic reactions depend on the strength of acids and bases.  ­ If a Strong acid is a reactant and a weak acid is a product, then the products are  favored.  ­% dissociation = [HA dissolved]/[HA initial]   Polyprotic acids: Acids with more than one ionizable proton.  ­ The dissociation of the first proton is easier than that of the second, which is  similarly easier than that of the third proton.  ­ Examples of polyprotic acids are H2SO4, and H3PO4.  ­ All strong polyprotic acids become weak when they lose their first H+.  The strength of an acid depends on 2 factors:  ­Strength of the HA bond. Stronger HA bonds = weaker acids.  ­HA bond polarity, electronegativity difference.                ­ Greater bond polarity = stronger acids.   Salts may affect pH of a solution in some situations:  ­ Salts must have cation or anion of a weak acid or base.  ­ Ex: NaCl in a solution becomes neutral because NaOH is a strong base and HCl is a strong acid, so they cancel each other out.  ­ NH4Cl is acidic because NH3 is a weak base and HCl is a strong acid, so acid  wins.  ­ ZnBr is acidic because HBr is a strong acid and ZnOH is a weak base.   Acid base buffers: Solution that has no pH change when a strong acid or base is added  due to weak acid/base and its conjugate pair.  ­Higher concentration buffers are more effective.   Buffers contain a common ion between the acid and the base, which causes the  neutralizing effect toward a strong acid or base’s addition to the solution.   Buffers work under Le Chat’s principle, so: ­Adding acid to a buffer causes the system to consume it and make HA. ­Adding base to a buffer causes the system to consume it and make A­.   For good examples of buffer problems see the posted exam 3 examples on Blackboard. 


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