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Chem 2 Heider > Ch 16 - Aq Ionic Equilibrium > Flashcards

Ch 16 - Aq Ionic Equilibrium Flashcards

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1
Q

buffer

A

a solution that resist pH change

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2
Q

solubility equilibria

A

the extent to which slightly soluble ionic compounds dissolve in water

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3
Q

an important blood buffer is

A

the mixture of carbonic acid(H2CO3) and the bicarbonate ion(HCO3-)

	- H2CO3(aq) + OH-(aq) -> H2O(l) + HCO3-(aq)
		- HCO3-(aq) + H+(aq) -> H2CO3(aq)
	- this combination keep the blood pH nearly constant
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4
Q

acidosis

A

a condition in which the acid affects the equilibrium between hemoglobin(Hb) and oxygen

	- HbH+(aq) + O2(g)  HbO2(aq) + H+(aq)
		- excess H+ causes the equilibrium to shift left reducing the bloods ability to carry oxygen resulting in hyperventalization with the potential to lead to coma or death
		- drinking antifreeze is bad
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5
Q

buffer

A

resist pH change by neutralizing added acid or added base

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6
Q

A buffer contains either:

A
  • significant amounts of a weak acid and its conjugate base or
    - significant amounts of a weak base and its conjugate acid
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7
Q

the buffer in blood is composed of

A

carbonic acid(H2CO3) and the conjugate base bicarbonate ion(HCO3-)

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8
Q

a weak acid or a weak base does not contain

A

enough of the opposite to be a buffer

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9
Q

a buffer must contain significant amounts of both

A

a weak acid and its conjugate base(or vice versa)

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10
Q

buffer characteristics

A
  • resist pH change
    - contains a significant amount of either 1) a weak acid and its conjugate bade or 2) a weak base and its conjugate acid
    - the weak acid neutralizes the added base
    - the base neutralizes the added acid
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11
Q

common ion effect

A

the presence of an ion results in a less acidic solution(higher pH) because adding the ion causes the equilibrium to shift left based on Le Chatliers principle
- the acid ionizes even less than normal

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12
Q

Henderson-Hasselbalch Equation

A

pH = pKa + log([base]/[acid])

		- holds true as long as the x is small approximation is valid
			- typically when initial concentrations are not too dilute AND equilibrium constant is fairly small
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13
Q

calculating the pH changes in buffer solutions requires two steps:

A
  • stoichiometry calculation

- equilibrium calculation

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14
Q

The Stoichiometry Calculation

A

calculate how the addition changes the relative amounts of acid and conjugate base

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15
Q

The Equilibrium Calculation

A

calculate the pH based on the new amounts of acid and conjugate base

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16
Q

The Stoichiometry Calculation:

as the acid is neutralized a stoichiometric amount of the base is converted into the conjugate acid through the neutralization reaction:

A
  • H+(aq) + A-(aq) -> HA(aq)
    - neutralizing .025 mol of H+ requires .025 mol of the weak base
    - H+ adds 0.025 mol while A- decreases by 0.025 and HA adds 0.025 mol
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17
Q

The Equilibrium Calculation:

figuring out the stoichiometric calculations provides the new initial concentrations which can be used to calculate the new pH

A 
-  [HA]		[H3O+]		[A-]
	I	0.125		0.00		0.075
	C	-x		+x		+x
	E	0.125-x		x		0.075+x
	-  x = [H3O+] = Ka(0.125/0.075)
		-  Ka = [H3O+][A-]/[HA]
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18
Q

When calculating the pH of a buffer after adding small amounts of acid or base remember:

A
  • adding a small amount of a strong acid to a buffer converts a stoichiometric amount of the base to the conjugate acid and decreases the pH of the buffer(adding acid decreases pH as expected)
    - adding a small amount of a strong base to a buffer converts a stoichiometric amount of the acid to the conjugate base and increases the pH of the buffer(adding base increase the pH as expected)
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19
Q

two factors influencing the effectiveness of a buffer

A
  • relative amounts of acid and base

- absolute concentrations of the acid and conjugate base

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20
Q

relative amounts of acid and base

A
  • a buffer is most effective(most resistant to pH change) if the concentrations of acid and conjugate base are equal
    - in order for a buffer to be reasonably effective, the relative concentrations of acid and conjugate base should not differ by more than a factor of 10
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21
Q

Absolute Concentrations of the acid and conjugate base

A
  • a buffer is most effective(most resistant to pH change) when the concentrations of acid and conjugate base are high
    • the buffer range is defined:
      • pH = pKa +- 1
        • A pH of 4 could have a buffer anywhere from pH 3-5
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22
Q

buffer capacity

A

the amount of acid or base you can add to a buffer without causing a large change in pH

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23
Q

buffer capacity increases with

A

increasing absolute concentrations of the buffer components

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24
Q

overall buffer capacity increases as

A

the relative concentrations of the buffer components become more similar to each other
- the closer to a 1:1 ration the more effective the buffer becomes

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25
Q

acid-base titration

A

a basic or acidic solution of an unknown concentration reacts with an acidic or basic solution of known concentration
- the know solution is slowly added to the unknown while the pH is monitored

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26
Q

indicator

A

a substance whose color depends on the pH

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27
Q

equivalence point

A

the point in the titration when the number of moles of base is stoichiometrically equal to the number of moles of acid
- the titration is complete here, neither reactant is in excess

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28
Q

titration or pH curve

A

a plot of the pH of the solution during a titration

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29
Q

The Titration of a Strong Acid with a Strong Base(base to acid in reverse of below):

Step 1

find volume NaOH required to reach the equivalence point

A
  • HCL = 0.0250L(0.100mol/1L) = 0.00250mol HCL
    - 0.00250mol(1L/0.100mol) = 0.0250L volume NaOH
    - equivalence reached at 25mL addition NaOH
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30
Q

The Titration of a Strong Acid with a Strong Base(base to acid in reverse of below):

Step 2

Initial pH(before adding any base)

A

pH = -log(0.100) = 1.00 HCL

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31
Q

The Titration of a Strong Acid with a Strong Base(base to acid in reverse of below):

Step 3

pH after adding 5.00mL NaOH

A
  • 0.00500L(0.100mol/1L) = 0.000500mol NaOH
    - [H3O+] = 0.00200mol[H3O+]/0.0250L + 0.00500L = 0.0667M
    - 0.002 = 0.00250 initial HCL – 0.0005mol change
    - final mol H3O+/initial volume(added volume)
    - pH = -log(0.0667) = 1.18
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32
Q

The Titration of a Strong Acid with a Strong Base(base to acid in reverse of below):

Step 4

pH after adding 10.0, 15.0, 20.0 mL NaOH

A

1.37, 1.60,1.95

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33
Q

The Titration of a Strong Acid with a Strong Base(base to acid in reverse of below):

Step 5

pH after adding 25.0 mL NaOH(equivalence point)

A

the pH of a strong acid-strong base titration will always be 7.00 at 25C.

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34
Q

The Titration of a Strong Acid with a Strong Base(base to acid in reverse of below):

Step 6

pH after adding 30.00mL NaOH

A
  • subtract the initial amount of H3O+ from the amount of OH- added
    - mol OH- added = 0.0300L(0.100mol/1L) = 0.00300mol OH-
    - [OH-] = 0.000500mol OH-/0.0250L + 0.0300L = 0.00909M
    - [H3O+] = 110^-14/0.00909 = 1.1010^-12
    - pH = -log(1.10*10^-12 = 11.96
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35
Q

The Titration of a Strong Acid with a Strong Base(base to acid in reverse of below):

Step 7

pH after adding 35.0,40.0,50.0 mL NaOH

A
  • 12.22,12.36,12.52
36
Q

The Titration of a Strong Acid with a Strong Base(base to acid in reverse of below):

Step 8

Overall pH curve

A

S shaped curve

starting at low pH

37
Q

Summarizing the Titration of a Strong Acid with a Strong Base:

the initial pH is simply the

A

pH of the strong acid solution to be titrated

38
Q

Summarizing the Titration of a Strong Acid with a Strong Base:

before the equivalence point

A

H3O+ is in excess. Calculate the [H3O+] by subtracting the number of moles added OH- from the initial number of moles H3O+ and dividing by the total volume

39
Q

Summarizing the Titration of a Strong Acid with a Strong Base:

at the equivalence point

A

neither reactant is in excess and the pH = 7.00

40
Q

Summarizing the Titration of a Strong Acid with a Strong Base:

beyond the equivalence point

A

OH- is in excess. Calculate the [OH-] by subtracting the initial number of moles H3O+ from the number of moles of added OH- and dividing by the total volume

41
Q

Summarizing Titration of a Weak Acid with a Strong Base:

the initial pH is that of the

A

weak acid solution to be titrated. Calculate the pH by working an equilibrium problem using the concentration of the weak acid as the initial concentration

42
Q

Summarizing Titration of a Weak Acid with a Strong Base:

between the initial pH and the equivalence point,

A

the solution becomes a buffer. Use the reaction stoichiometry to calculate the amounts of each buffer component and then use the Henderson-Hasselbach equation to calculate the pH

43
Q

Summarizing Titration of a Weak Acid with a Strong Base:

Halfway to the equivalence point

A

the buffer components are exactly equal and pH-pKa

44
Q

Summarizing Titration of a Weak Acid with a Strong Base:

at the equivalence point

A

basic at equivalence point

the acid has all been converted into its conjugate base. Calculate the pH by working an equilibrium problem for the ionization of water by the ion acting as a weak base.

  • calculate the concentration of the ion acting as a weak base by dividing the number of moles of the ion by the total volume at the equivalence point
45
Q

Summarizing Titration of a Weak Acid with a Strong Base:

Beyond the equivalence point

A

OH- is in excess. Ignore the weak base and calculate the [OH-] by subtracting the initial number of moles of the weak acid from the number of moles of added OH- and dividing by the total volume.

46
Q

when a polyprotic acid has a Ka1 and Ka2 sufficiently different then

A

the curve will have two equivalency points

47
Q

indicators allow

A

the monitoring of a titrations pH

48
Q

endpoint

A

the point where the indicator color changes to determine the equivalency point

49
Q

the correct indicator will endpoint of the titration will

A

trigger a color change

50
Q

an indicator is itself a weak organic acid that is

A

is a different color than its conjugate base

51
Q

Indicator color summary

A
  • pH = pKa then ratio = 1 and color is the intermediate color
    • pH = pKa + 1 then ratio = 10 and color is the base
    • pH = pKa – 1 then ratio = 0.10 and color is the acid
    • only usable within +-1 or a magnitude of 10 before the indicator is out of range
52
Q

solubility product constant(Ksp)

A

the equilibrium constant for a chemical equation representing the dissolution of an ionic compound

53
Q

molar solubility

A

the solubility is units of moles per liter(mol/L)

- NOT the Ksp but the solubility product constant

54
Q

the solubility product constant(Ksp) will have the

A

same value at a specific temperature

55
Q

the solubility can have different values in

A

different kinds of solutions

56
Q

Solubility example

AgCl(s) Ag+(aq) + Cl-(aq)

A 
I			0.00	0.00
	C			+S	+S
	E			S	S
		-  S represents the concentration of AgCl that dissolves(molar solubility)
		-  Ksp = [Ag+][Cl-]
			  =S^2
			S = sqrtKsp
			= sqrt(1.77*10^-10)
			S = 1.33 * 10^-5 M or AgCl molar solubility is 1.33*10^-5 mol/L
57
Q

the Ksp value can only be compared directly as a measure of relative solubility if

A

the items being compared have the same dissociation stoichiometry(both have 1mol of each compound produces 3 mol of dissolved ions is =)

58
Q

in general, the solubility of an ionic compound is lower in

A

a solution containing a common ion than in pure water

  • Le Chatliers principle at work, more products shifts equilibrium left and less soluble
59
Q

in general, the solubility of an ionic compound with a strongly basic or weakly basic anion increases with

A 
increasing acidity(decreasing pH)
		-  low pH more soluble = equilibrium shift right making more OH- and vice versa
60
Q

common basic anions include:

A
  • OH- Hydroxides
    - S^2- sulfides
    - CO3^2- carbonates
    - all are more soluble in acidic water than in pure water
    - this is why rainwater dissolve rocks high in limestone(CaCO3)
61
Q

Q<Ksp

A

the solution is unsaturated and more of the solid ionic compound can dissolve in the solution

62
Q

Q=Ksp

A

the solution is saturated, the solution is holding the equilibrium amount of the dissolved ions and additional solid does not dissolve in the solution

63
Q

Q>Ksp

A

the solution is supersaturated, under most circumstances the excess solid precipitates out of a supersaturated solution

64
Q

selective precipitation

A

process involving the addition of a reagent that forms a precipitate with one of the dissolved cations but not the others

65
Q

qualitative analysis

A

involves finding the kind of ions present in the solution

66
Q

quantitative analysis

A

involves finding the amounts of substances in a solution or mixture

67
Q

5 groups of precipitates

A
  1. Insoluble Chlorides
  2. Acid-insoluble Sulfides
  3. Base-Insoluble Sulfides and Hydroxides
  4. Insoluble Phosphates
  5. Alkali Metal Ions and Na+, K+, and NH4+
68
Q

5 groups of precipitates:

Insoluble Chlorides

A

AgCl, Hg2Cl2, PbCl2

69
Q

5 groups of precipitates:

Acid-insoluble sulfides

A

CuS, Bi2S3, CdS, PbS, HgS, As2S3, Sb2S3, SnS2

70
Q

5 groups of precipitates:

base-insoluble sulfides and hydroxides

A

Al(OH)3, Fe(OH)3, Cr(OH)3, ZnS, NiS, CoS, MnS, FeS

71
Q

5 groups of precipitates:

insoluble phosphates

A

Ba3(PO4)2, Ca3(PO4)2, MgNH4PO4

72
Q

5 groups of precipitates:

alkali metal ions and Na+, K+, NH4+

A

alkali metal ions and Na+, K+, NH4+

73
Q

Group 1: Insoluble Chlorides

A

Ag+, Pb2+, and Hg2^2+ precipitate out

74
Q

Group 2: Acid-Insoluble Sulfides

A

at low pH the equilibria shifts left and at high it shifts right
- results in S2- being unavailable at low pH but at high pH S2- is able to precipitate

75
Q

Group 3: Base-Insoluble Sulfides and Hydroxides

A

the added base reacts with acid and after the acid insoluble sulfides have precipitated to precipitate out

76
Q

Group 4: Insoluble Phosphates

A

alkaline earth metal cations can be precipitated by adding (NH4)HPO4 causing metal phosphates to form

77
Q

Group 5: Alkali Metals and NH4+

A

Na+, K+, and NH4+

	- they can not be precipitated but presence can be determined
		- Na+(yellow/orange flame) and K+(blue flame) by flame test
78
Q

complex ion

A

contains a central metal ion bound to one or more ligands

79
Q

ligand

A

a neutral molecule or ion that acts as a Lewis base with the central metal ion

80
Q

formation constant(Kf)

A

the equilibrium constant associated with the reaction for the formation of a complex ion

81
Q

in general the values of Kf are

A

very large indicating that the formation of complex ions is highly favored

82
Q

the solubility of an ionic compound containing a metal cation that forms complex ions increases in

A

the presence of Lewis bases that complex the cation

83
Q

most common Lewis bases that increase the solubility of metal cations are:

A

NH3, CN-, and OH-

84
Q

metal hydroxides are highly insoluble in pH-neutral water but

A

but become highly soluble in acidic solutions

- Al(OH)3 acts as a base for acidic solutions by reacting with the H3O+

85
Q

some metal hydroxides can also act as acids

A
  • amphoteric
    - Al(OH)3(s) + OH-(aq) -> Al(OH)4-(aq)
    - accepts H+ and acts as an acid
    - soluble in high and low pH BUT insoluble in a pH neutral solution
86
Q

metal hydroxide cations that are amphoteric:

A

Al^3+, Cr^3+,Zn^2+,Pb^2+, Sn^2+

Chem 2 Heider (9 decks)

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