# Chem 1220 Midterm 2

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1. In equilibrium, there is no net change of...
Concentration (Vapor Pressure)
2. Equilibrium (liquid-vapor phase) change in volume
If volume is increased, pressure falls, more gas vaporizes, and pressure is restored.

If volume is decreased, pressure rises, more gas condenses, and pressure is restored.
3. Chemical Equilibrium
The state where the concentrations of all reactants and products remain constant with time.
4. K= ?
The equilibrium rate constant. It relates the concentrations of reactants. and products in equilibrium.
5. K controls...
the ratio of concentrations... not the other way around.
6. What happens to K if...
Double the stoichiometry?
Reverse the order the equilibrium equation is written?
K-new= (K-original)^n  in this case 2

K-new= 1/K-original
7. K vs temperature
K always has the same value at a given temperature regardless of the amounts of reactants or products are present initially.
8. Do the equilibrium concentrations in solution change because we write the equation multiplied by a factor?
No, the chemicals do not know or care how we write the raction... if we double the reaction equation we have the same concentrations, but their quotient is squared to give the new constant K2=k1^2
9. Kp
• equilibrium constant for partial pressures.
• Kp=K(RT)^(delta)n

(delta)n=sum of coefficients of the gaseous products minus the sume of the coefficients of the gaseous reactants.

R=0.08206 L atm/mol K

T= temperature (in kelvin)
10. Homogeneous equilibria
all the components are in the same phase
11. Heterogeneous equilibria
the components are in more than one phase
12. Reaction Quotient Q
same expression as the equilibrium constant K, but does not assume the concentrations are in equilibrium.
13. Q vs K
Q indicates in which direction the reaction will proceed.

If Q<K then the system has too much reactant and will shift to the right.

If Q=K then the system is in equilibrium

If Q>K then the system has too much product and will shift to the left.
14. Solving Equilibrium Problems
1) write the balanced equation for the reaction

2) write the equilibrium expression

3) List the initial concentrations

4)Calculate Q, and determine in which direction the concentrations will change.

5) Define the change needed to reach equilibrium and define the equilibrium concentrations by applying the change to the initial concentrations.

6) Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown.

7) Check your calculated equilibrium concentrations by making sure they give the correct value of K.
15. Le Chatelier's Principle
if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to counteract that change.

note that the position of equilibrium (the concentrations of reactants and products) shift, but the reaction constant K does not change if the temperature stays constant.
16. Le Chatelier's Principle and heat
treat the addition or removal of heat the same as you would a quantity of reactants/products
17. Le Chatelier's Principle  and pressure
When the pressure is increased, the reaction moves towards decreasing the pressure: towards the side that has lower number of moles of gas.
18. Le Chatelier's Principle and volume
the constant of the reaction depends on concentration or pressures, not volumes, so we have to evaluate how the effect of volume affects the pressures or concentrations.
19. Le Chatelier's Principle and inert gas
addition of an inert gas does not affect the equilibrium position.
20. Le Chatelier's principle and temperature
Treat the addition or removal of heat the same as you would a quantity of reactants/products.
21. Arrhenius definition of acids and bases
acid= a substance that produces H+ ions in solution

base= a substance that produces OH- ions in solution.
22. Bronsted-Lowry definition of acids and bases
Acids are proton (H+) donors

Bases are proton acceptors
23. Lewis definition of acids and bases
Acids are electron pair acceptors

Bases are electron pair donors.
24. Conjugate acid and base
conjugate base- is everything that remains of the acid molecule after a proton is lost

conjugate acid- is formed when the proton is transferred to the base.
25. Strong acid vs weak acid
• Strong acid
•     Ionization equilibrium lies far to the right
•     Yields a weak conjugate base.
26. Acids and bases  and organic molecules
an organic molecule that contains a carboxylic group is acidic.  The H of the carboxylic group detaches as H+ in water, leaving a carboxylate R-COO-
27. Some properties of acid solutions
• Have sour taste.
• Change the color of acid/base indicators.
• React with bases to produce salts and water.
• React with salts of a weaker acid to form the weaker acid and a salt of the stronger acid.
• React with metals to produce H2 (not an acid/base reaction.
• Conduct electricity (because they produce ions)
28. Some properties of basic solutions
• Produce a slippery feel
• Have a bitter taste
• Change the color of indicators
• Conduct electricity (they produce ions)
• Combine with acids to form salts plus water.
29. Major differences between Arrhenius and Bronsted-Lowry theories
1. the reaction does not have to occur in an aqueous solution. (Bronsted-Lowry)

2. bases are not required to be hydroxides. (Bronsted-Lowry)
30. Acidity constant Ka
is defined for the reaction of an acid when water acts as the base.
31. Magnitude of the acidity constant tells us...
how strong an acid is
32. Kw
the dissociation constant of water, it tells that the product of [H+] and [OH-] is the same in all aqueous solutions

Kw= [H+][OH-] = 10^-14
33. amphoteric substance
substance that behaves like an acid or like a base
34. [H+]=[OH-]
neutral solution
35. [H+] > [OH-]
acidic solution
36. [H+] < [OH-]
basic solution
37. Kb
basicity constant= reaction of base to take H+

Kb = ( [OH-][HA] ) / [A-]
38. Ka
Ka = ( [H+][A-] ) / [HA]
39. Degree of ionization
%ionization = H+ produced / HA original x 100%
40. Kw Ka and Kb related
Kw = Ka x Kb

valid only if the acid and base are conjugate
41. pH
-log[H+]

way to represent solution acidity

pH decreases as [H+] increases

pH=7; neutral

pH > 7; basic

pH < 7; acidic
42. Going from pH to proton concentration
10^-[H+]
43. pKw
pH + pOH

pKw = 14
44. How to approach acid-base problems
• 1. Find major species in solution
• 2. find dominant reaction that will take place
•      a)is it an equilibrium reaction or a reaction         that will go essentially to completion?             If it is strong, it is a complete reaction.        b)React all major species until you are left         with an equilibrium reaction.
• 3. solve for pH if needed.
45. What is the pH of a very very dilute solution of an acid?
There is less H+ from the acid than from dissociation of water; so the dominant reaction is H20 + H2O --> H3O+ + OH-

pH is 7... very dilute solutions of acids are slightly acidic or neutral... but never are basic.
46. Calculating the pH of weak acid solutions
• 1. List the major species in the solution
• 2. choose the species that can produce H+, and write balanced equations for the reactions producing H+
• 3. using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H+.
• 4. Write the equilibrium expression for the dominant equilibrium.
• 5.List the initial concentrations of the species participating in the dominant equilibrium.
• 6. Define the change needed to achieve equilibrium; that is, define x.
• 7. write the equilibrium concentrations in terms of x
• 8. substitute the equilibrium concentrations into the equilibrium expression.

note: x is negligible when subtracting from weak acid. But make sure to check that it is less than 5%
47. Percent Dissociation (Ionization)
percent dissociation = (amount dissociated (mol/L)) / (initial concentration (mol/L)) x 100%

For a given weak acid, the percent dissociation increases as the acid becomes more dilute.
48. Bases
pH calculations for solutions of weak bases are very similar to those for weak acids.

Kw= [H+][OH-] = 1.0 x 10^-14

pOH= -log[OH-]

pH= 14.00 - pOH
49. Polyprotic Acids
Acids that can furnish more than one proton

Always dissociates in a stepwise manner, one proton at a time.

The conjugate base of the first dissociation equilibrium becomes the acid in the second step.

• For a typical weak polyprotic acid:
•     Ka1 > Ka2 > Ka3

For a typical polyprotic acid in water, only the first dissociation step is important to pH.
50. Acid-Base properties of salts
salts:

ionic compounds

when dissolved in water, break up into its ions (which can behave as acids or bases)

The salt of a strong acid and a strong base gives a neutral solution

A basic solution is formed if the anion of the salt is the conjugate base of a weak acid.

An acidic solution is formed if the cation of the salt is the conjugate acid of a weak base.
51. The Effect of structure on Acid-Base properties
Factors for acidity in binary compounds:

Bond polarity: high bond polarity makes the bond easier to break.

Bond strength: low bond strength makes the bond easier to break

Stability and hydration of the anion: the more stable the anion, the more it will dissociate.
52. Buffered Solutions
Mixtures with relatively high and comparable concentrations of acid and its conjugate base (or base and its conjugate acid) are good at controlling (buffering) the pH of a solution.

• Buffered solutions- resist change in pH
• They are weak acids or bases w/ common ion
• after addition o stron acid or base, deal with stoichiometry first, then the equilibrium.
53. pH of a solution to the ratio of acid and conjugate base in buffers.
Henderson-Hasselbach equation

pH= pKa + log ([A-] / [HA])

This equation is always valid

It tells us that if the equilibrium concentration of acid and conjugate base are the same, pH=pKa

It is also practical to compute the pH of buffers.
54. Buffering Capacity
Buffering capacity indicates the amount of protons or hydroxide ions the buffer can absorb without a significant change in pH.
55. How do you maximize the buffering capacity of a particular buffer?
Use about the same amount of acid and conjugate base. and use a large number of moles of both acid and its conjugate base.
56. Titration
quantitative neutralization of an acid or a base. It is employed to quantify the amount of acid or base.
57. Equivalence point
point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated.
58. Solubility product
Ksp=Keq

The equilibrium constant for the dissociation of a solid salt into its aqueous ions.
59. Solubility (S)
an equilibrium position

Also known as molar solubility

Solubility is the number of moles of salt that will dissolve in a liter of solution (At a given T)

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 Author: Anonymous ID: 299484 Filename: Chem 1220 Midterm 2 Updated: 2015-03-30 03:19:30 Tags: chem 1220 midterm2 Folders: Description: Chem 1220 Midterm 2 Show Answers:

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