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MCAT > Chem I: 7-10 > Flashcards

Chem I: 7-10 Flashcards

1
Q

titration

use

A

used to determine the concentration of known reactant in a solution

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2
Q

how are titration performed?

A

adding small volumes of known concentration (titrant) to a known volume of a solution of unknown concentration (titrand) until completion of the reaction is achieved at the equivalence point

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3
Q

acid base titrations

equivalence point

A
  • reached when the number of acid equivalents present in the original solution equals the number of base equivalents, or vice versa
  • steepest point of titration curve
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4
Q

titrant

A

solution of known concentration that is added

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5
Q

titrand

A

solution of known volume but unknown concentration

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6
Q

equation that allows us to calculate unknown concentration of titrand

A

NaVa = NbVb

N = normalities

V = volume (as long as volumes use same units, do not have to be liters)

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7
Q

two ways the equivalence point in acid-base titration is calculated:

A
  1. graphical methods: plotting pH of unknown solution as a function of added titrant by using pH meter
    1. midpoint of the region of the curve with steepest slope
  2. estimated by watching for a color change of an added indicator
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8
Q

indicator

A

weak organic acids or bases that have different colors in pronated and deprotonated states (conjugate acid and conjugate base forms) –> the binding or release of a proton leads to change in absorption spectrum of molecule

bc of le chatelier’s principle

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9
Q

titration

indicator must always…

A

be weaker acid or base than the acid or base being titrated

other wise the indicator would be titrated first

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10
Q

endpoint

A

point at which the indicator changes to its final color

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11
Q

accuracy of weak acid/weak base titration

A

not accurate

  • titration lacks the sharp change that normally indicates equivalence point
  • indicators are less useful because the pH change is far more gradual
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12
Q

steps for questions involving the selection of an ideal indicator

A
  • determine where equivalence point is
  • select the indicator that has the closest pKa value to it
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13
Q

if strong acid + weak base, equivalence point…

A

equivalence point pH < 7

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14
Q

if strong acid + strong base, equivalence point…

A

pH = 7

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15
Q

if weak acid + strong base, equivalence point…

A

pH > 7

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16
Q

strong acid and strong base titration

A
  • early part of curve: little base added -> acidic species predominates
    • addtn of small amounts of base will not appreciably change the OH- or pH
  • equivalence point
    • adding base will elicit the most substantial changes in pH
  • last part of curve: excess base added -> small amounts of base will not change OH- or pH significantly
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17
Q

weak acid and strong base

titration

A
  • inital pH in acidic range
  • pH changes gradually early on intitration
    • has a less sudden rise at equivalence point
  • equiv point pH > 7
    • basic range
  • reaction produces weak conjugate base (A-) and weaker conjugate acid (H2O)
    • produces greater conc of OH- than H+ at equilibrium (due to common ion effect on autoionization of water)
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18
Q

strong acid and weak base

titration

A
  • inital pH in basic range
  • graduate drop in pH with addition of strong acid
  • equiv point pH < 7
    • acidic range
  • will produce a weak conjugate acid and weaker conjugate base
    • higher concentration of H+ ions
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22
Q

weak acid and weak base

titration

A
  • inital pH generally in 3-11 range
  • very shallow drop at equiv point
  • equiv point will very near neutral pH because the reaction is partially dissociative for both species
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23
Q

titrations

the stronger the acid or base….

A

the more it pulls the equivalence point into its pH territory

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24
Q

to identify which type of titration is being shown in a graph, identify starting position

pH >> 7

A

titrand is strong base

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25
Q

polyvalent titration

A
  • multiple equivalence points
  • first buffer region: flat part of curve
  • half equivalence pt: center of buffer region (pt between region I and II)
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26
Q

to identify which type of titration is being shown in a graph, identify starting position

pH > 7

A

titrand is weak base

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27
Q

to identify which type of titration is being shown in a graph, identify starting position

pH < 7

A

titrand is weak acid

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28
Q

to identify which type of titration is being shown in a graph, identify starting position

pH << 7

A

titrand is strong acid

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30
Q

half equivalence point

A

occurs when half a given species has been protonated or deprotonated

when pH = pKa, pOH = pKb

buffering capacity is optimal

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31
Q

titrations of acidic and basic amino acids

A
  • similar to polyvlaent titration
  • 3 equivalence points
    • one corresponding to titration of carboxyl group, amino group, acidic or basic side chain
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32
Q

pKa =

pKb =

A

pKa = -log [Ka]

pKb = -log [Kb]

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33
Q

Ka =

A

Ka = ( [H+][A-] ) / [HA]

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34
Q

Kb =

A

Kb = ( [HA][OH-] ) / [A-]

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35
Q

weak acid/base

steps

A

get rid of x and approximate

ICEbox - LR

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36
Q

KaKb = [?]

A

KaKb = [H3O+][OH-]

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37
Q

pH =

pOH =

A

pH = -log[H+]

pOH = -log[OH-]

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38
Q

autoionization of water

A

H2O = H+(aq) + OH-(aq)

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39
Q

pKw =

A

pKw​ = pH + pOH = 14

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40
Q

pH = pKa when

A

midpoint of titration of weak acid and strong base

[HA] = [A-]

equal amounts of conjugate acid and base

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41
Q

pH = pKa when

A

midpoint of titration of weak acid and strong base

[HA] = [A-]

equal amounts of conjugate acid and base

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42
Q

buffer

A

weak acid and weak conjugate base (salt), or weak base and its salt

suppresses the change in pH when small amounts of acid or base are added

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43
Q

Henderson Hasselbach eq

A

buffer problem

  • Ka for conjugate acid base pair: 10-4 to 10-11 M
  • ratio base/acid: 0.1-10
  • values of [base] and [acid]: 10-3 to 1 M
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44
Q

clues identifying reaction type, steps to solve for pH

strong acid and strong base

A
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45
Q

clues identifying reaction type, steps to solve for pH

weak acid alone

A
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46
Q

clues identifying reaction type, steps to solve for pH

weak base alone

A
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47
Q

clues identifying reaction type, steps to solve for pH

weak acid and strong base

equal moles of both

A
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48
Q

clues identifying reaction type, steps to solve for pH

weak acid and strong base

more moles of weak acid than strong base

A
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49
Q

clues identifying reaction type, steps to solve for pH

weak base and strong acid

equal moles for both

A
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50
Q

clues identifying reaction type, steps to solve for pH

weak base and strong acid

more moles of weak base

A
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51
Q

clues identifying reaction type, steps to solve for pH

buffer

A
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52
Q

clues identifying reaction type, steps to solve for pH

perturbed buffer

A
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53
Q

rules

strong acids and strong bases

calculating effect on pH

A
  • dissociate completely.
  • calculating effect on pH: determine [H+] or [OH-] by determining moles of acid or base per L and recalling that every strong acid and strong base will completely dissociate
  • concentration of acid or base is equal to H+ or OH-
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54
Q

rules

weak acids and weak bases

A
  • dissociate or associate only partially
    • degree to which they dissociate (for acids) or associate (for bases) is determined by the equilibrium constant for their characteristic acid or base reaction (Ka or Kb)
  • pKa = -log Ka and pKb = -log Kb
  • pKa is also equal to the pH at which an acid is half protonated and half deprotonated
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55
Q

neutralization

final pH w strong acid and strong base added together

A
  • strong acids and strong bases when put together in solution strive to neutralize each other
    • ex: H+ + OH- –> H2O K = 1 x 1014
  • To determine the final pH when a strong acid and strong base are added together, one must figure out if either H+ or OH- is limiting and then determine how much OH- or H+ is left over after the neutralization has occurred and calculate the pH from that.
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56
Q

to calculate pH of buffer

A

use the Henderson-Hasselbach equation

(works only when conjugate acid and conjugate base are present)

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57
Q

Within how many pH units of a buffer’s pKa is the buffering capacity said to be maintained?

(A) 1 unit
(B) 2.5 units
(C) 0.5 units
(D) This depends on the buffer

A

A) 1 unit

Within 1 pH unit of its pKa, a buffer is said to maintain its buffering capacity (have an ability to resist changes in pH).

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58
Q

How do the titration curves of a strong acid and strong base compare to a weak acid and a strong base? Consider buffering capacity, the shapes of the curves, and the equivalence points.

A

For strong acid, notice the little change in pH until right before the equivalence point, then a sharp increase.
For the weak acid, notice the gradual change in pH as base is added, showing buffering capacity. The middle of that buffering capacity is where pH=pKa. Near its equivalence point, it will also sharply increase.
The equivalence points are halfway between the pH before and after the steep increase.

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62
Q

a) the end point and equivalence will both occur later than expected
b) the end point will now occur significantly before the equivalence point is reached
c) the end point will now occur significantly after the equivalence point
d) the end point and equivalence point will both occur sooner than expected

A

c) the end point will now occur significantly after the equivalence point

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63
Q

a) this is a suitable indicator, and it will change color after the equivalence point is reached
b) the solution will not change color until well after the equivalence point is reached
c) when the first drop of titrant is added, the solution will change color
d) the indicator will react with the analyte immediately, changing color without the titrant being added

A

d) the indicator will react with the analyte immediately, changing color without the titrant being added

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64
Q

a) 2:1, because after the conjugate base reacts with the weak acid, the dynamic range of buffering is equally split between preventing changes caused by acids and bases
b) 1:1, because the dynamic range of buffering is equally split between preventing changes caused by acids and bases
c) 2:1, because the dynamic range is larger and covers the widest range of pH’s
d) 1:1, because the dynamic range is larger and covers the widest range of pH’s

A

b) 1:1, because the dynamic range of buffering is equally split between preventing changes caused by acids and bases

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65
Q

a) tripling the amount of both acid and base added to the solution
b) adding salts to supplement the buffering effects
c) adding more base to increase the dynamic range of the solution
d) removing salts to enhance the buffering effects

A

a) tripling the amount of both acid and base added to the solution

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66
Q

a) there will be no reaction, and the titrant must be replaced with the solution labeled “analyte”
b) the solution’s pH will change, and the titrant must be replaced with the solution labeled “analyte”
c) there will be no reaction, and the titrand must be replaced with the solution labeled “analyte”
d) the solution’s pH will change, and the titrand must be replaced with the solution labeled “analyte”

A

c) there will be no reaction, and the titrand must be replaced with the solution labeled “analyte”

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67
Q

a) 50.3 mL HCl, 49.7 NaOH
b) 39.5 mL HCl, 60.5 mL NaOH
c) 10 mL HCl, 90 mL NaOH
d) 57.4 mL HCl, 43.6 mL NaOH

A

a) 50.3 mL HCl, 49.7 NaOH

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68
Q

a) 133 mL
b) 100 mL
c) 50 mL
d) 67 mL

A

d) 67 mL

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76
Q

True or false? At the equivalence point, there is no buffering capacity because the titrand (what the titrant is being added to) has fully reacted with the titrant.

A

True. At the equivalence point, there is no buffering capacity because the titrand (what the titrant is being added to) has fully reacted with the titrant.

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77
Q

Which of the following is a typical range from which the indicator will have a color change?

(A) pKa ± 10
(B) pKa ± 0.1
(C) pKa ± 3
(D) pKa ± 1

A

(D) pKa + or - 1

Generally, an indicator (like phenolphthalein) will have its color change within 1 pH unit of its pKa. This can vary based on the indicator though!

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78
Q

You have a solution of 453 mL of HCl. If it takes 46.3 mL of .287 M NaOH to reach the end point of our titration of the HCl solution, what is the concentration of HCl in the original solution?

(A) .00342 M
(B) .00752 M
(C) .0132 M
(D) .0294 M

A

(D) .0294 M

46.3 mL NaOH ⋅ 1 L / 1000 mL ⋅ .287 moles NaOH / 1 L NaOH ⋅ 1 mole HCl / 1 mole NaOH = approx. .015 moles HCl (actual: .0133)

NOTE: You can use the M1V1 = M2V2 equation here, but I prefer to simply use dimensional analysis.

.015 moles HCl / .453 L HCl = approx. .03 M (actual: .0294 M)

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79
Q

You have a solution of 78.3 mL of HCl. If it takes 56.2 mL of .143 M Ba(OH)2 to reach the end point of our titration of the HCl solution, what is the concentration of HCl in the original solution?

(A) .0205 M
(B) .205 M
(C) 2.05 M
(D) 20.5 M

A

(B) .205 M

56.2 mL Ba(OH)2 ⋅ 1 L / 1000 mL ⋅ .143 moles Ba(OH)2 / 1 L Ba(OH)2 ⋅ 2 moles OH- / 1 mole Ba(OH)2 ⋅ 1 mole HCl / 1 mole OH- = approx. .015 moles HCl (actual: .01607)

.015 moles HCl / .0783 L HCl = approx. .2 M (actual: .205 M)

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82
Q

acid base nomenclature

acid s formed from anions with names that end in -ide

A

hydro- and -ic

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83
Q

At which of the following points is the pH equal to the pKa of the titrand?

(A) Equivalence point
(B) Endpoint
(C) Half-equivalence point
(D) Half-endpoint

A

(C) Half-equivalence point

Recall the Henderson-Hasselbalch equation and the nature of logarithms, especially that log(1)=0.

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84
Q

What is the pH of a 4.3 ⋅ 10^-4 M solution of 456.2 mL HCl after having added 23.4 mL of 5.6 ⋅ 10^-5 M NaOH?

(A) .67
(B) 1.43
(C) 2.32
(D) 3.39

A

(D) 3.39

  1. 4 mL NaOH ⋅ 1 L NaOH / 1000 mL NaOH ⋅ 5.6 ⋅ 10^-5 mol NaOH / 1 L NaOH = Approx. 10 ⋅ 10^-7 mol NaOH (actual: 13.1 ⋅ 10^-7)
  2. 2 mL HCl ⋅ 1 L HCl / 1000 mL HCl ⋅ 4.3 ⋅ 10^-4 moles HCl / 1 L HCl = Approx. 2000 ⋅ 10^-7 moles HCl (actual: 1961.66 ⋅ 10^-7)

2000 ⋅ 10^-7 moles HCl - 10 ⋅ 10^-7 mol NaOH = Approx. 1990 ⋅ 10^-7 mol HCl (actual: 1948 ⋅ 10^-7)

1990 ⋅ 10^-7 mol HCl / .4796 L = approx. 4000 ⋅ 10^-7 M (actual: 4062 ⋅ 10^-7)

pH = -log([H+])
pH = -log(4 ⋅ 10^-4)
pH = 3.5 (actual: 3.39)
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85
Q

acid base nomenclature

acids formed from oxyanions ending in -ite

A

(less oxygen)

-ous acid

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86
Q

acid base nomenclature

acids formed from oxyanions ending in -ate

A

(more oxygen)

-ic acid

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87
Q

units avagadro’s number

A

atoms/mol

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95
Q

arrhenius acid

A

dissociates to form excess of H+ in solution

contain H at beginning of formula

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96
Q

arrhenius base

A

dissociates to form excess of OH- in solution

contain OH at end of formula

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97
Q

bronsted lowry acid

A

donates H+ ions

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98
Q

bronsted lowry base

A

accepts H+ ions

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99
Q

calculate pH of 1 x 10-8 M solution of HCl

A
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100
Q

what is the pH of a solution with [HClO4] = 10 M?

A
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101
Q

advantage of bronsted lowry def of acid/base over arrhenius

A

not limited to solutions

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102
Q

every arrhenius acid/base can be classified as __A__ acid/base. every ___A___ acid/base can also be classified as ___B___.

does it work other way around?

A

bronsted lowry

lewis acid/base

not always

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103
Q

lewis acid

A

electon pair accpetor

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104
Q

lewis base

A

electron pair donor

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105
Q

which acid base def is most inclusive

A

Lewis

every Arrhenius acid is also a Bronsted-Lowry acid, and every Bronsted Lowry acid is also a Lewis acid (and likewise for bases).

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106
Q

amphoteric

A

species that reacts like an acid in a basic environment an like a base in acidic environment

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107
Q

amphiprotic

A

amphoteric species can either gain or lose a proton

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108
Q

things considered amphoteric and amphiprotic

A

water, amino acids, and partially deprotonated polyprotic acids (such as bicarbonate and bisuflate)

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109
Q

things considered amphoteric

why not amphiprotic

A
  • metal oxides and hydroxides
    • bc do not give off protons
  • partially dissociated conjugate base of polyvalent acid
  • species that can act as both oxidizing and reducing agents
  • amino acids that have zwitterion intermediates with both cationic and anion character
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111
Q

water reacting w base eq

A

H2O + B- ⇄ HB + OH-

112
Q

water reaction with acid eq

A

HA + H2O ⇄ H3O+ + A-

115
Q

write acid formula and acid name

anion: MnO4-

A

HMnO4

permanganic acid

116
Q

write acid formula and acid name

anion: I-

A

HI

hydroiodic acid

117
Q

what happens when strong acid and weak base react

A

cation of a salt is a weak acid and will react with water solvent

eventually makes slightly acidic solution

118
Q

what happens when strong base and weak acid react

A

anion will react with water, making slightly basic solution

119
Q
A
120
Q
A
121
Q
A
122
Q
A
123
Q

Kw =

A

Kw = [H+][OH-] = 10-14

124
Q

Determine the conc of H+ and pH of solution of 0.2 M acetic acid.

Ka = 1.8 x 10-5

A
125
Q

if a species donates hydrogen ions to pure water…

A

the hydrogen ion conc will increase, causing the system to shift toward reactants in autoionization proces

result is decrease in OH ion conc and return to equilibrium state

126
Q

if a species that accepts H+ ions is added to pure water…

A

results in decrease in H+ conc

causes system to shift toward products, returning system to equilibrium

127
Q

Kw is dependent on…

and not…

A

temperature (standard is 25 deg C)

NOT conc, pressure, volume

128
Q

pH + pOH =

A

14

129
Q

Name the 7 Strong Acids

A
130
Q

Strong bases are formed from which two groups of the periodic table mixed with OH?

A

Groups 1A and 2A

131
Q

acidic pH

A

< 7

132
Q

basic pH

A

> 7

133
Q

converting Ka to pH

A

n x 10-m

p value = m - 0.n

134
Q

The Ka of acetic acid is 1.8 x 10-5. What is the pKa of acetic acid?

a) 8.4
b) 3.5
c) 6.2
d) 4.7

A

d) 4.7

135
Q

What is the pH of a 8 x 10-6 Ba(OH)2 solution?

a) 9.2
b) 6.6
c) 4.8
d) 13.4

A

a) 9.2

136
Q

Which of the following metals is not a strong base as a salt with OH-?

a) Ca
b) Li
c) Rb
d) Mg

A

d) Mg

137
Q

a) HI
b) HNO3
c) H2SO4
d) HClO2

A

d) HClO2

138
Q

a researcher prepares a solution with an OH- conc of 10-8 M by adding a week electrolyte. which of the following did the researcher likely add to the solution?

a) H2CO3
b) NaOH
c) NH3
d) HCl

A

a) H2CO3

139
Q

Stomach acid has a pH between 1.5 and 3.5 What is the OH- conc when the H3O+ conc is 5 x 10-4?

a) 2 x 10^-11
b) 5 x 10^-4
c) 2 x 10^-18
d) 5 x 10^-8

A

a) 2 x 10^-11

140
Q

a) I only
b) II only
c) I and III only
d) I, II, and III

A

c) I and III only

141
Q

a molecule that is amphoteric can act as both a base and an acid. which of the following is not amphoteric?

a) HSO4-
b) H2O
c) glycine
d) H2CO3

A

d) H2CO3

142
Q

Strong and weak acids both dissociate in water. How can strong and weak acids be quantitatively distinguished?

a) acids with a pKa value greater than 1 are strong acids
b) acids with Ka value of less than 0 are strong acids
c) acids with a pKa greater than 0 are strong acids
d) acids with a Ka value less than 1 are weak acids

A

d) acids with a Ka value less than 1 are weak acids

143
Q

Chloride ions (Cl-) and acetate ions (CH3COO-) are both conjugate bases of acids. Which is the weaker base?

a) chloride, bc it is the conjugate base of a weak acid
b) acetate, bc it is the conjugate base of a strong acid
c) acetate, bc it is the conjugate base of a weak acid
d) chloride, bc it is the conjugate base of a strong acid

A

d) chloride, bc it is the conjugate base of a strong acid

144
Q

a) [H3O+][A-] / [HA]
b) acid dissociation constant
c) pH
d) acid ionization constant

A

c) pH

145
Q

H2O and BF3 can react to form BF3OH2. For this reaction BF3 is considered a:

a) lewis acid
b) lewis base
c) leaving group
d) nucleophile

A

a) lewis acid

146
Q

a) bronsted lowry base
b) bronsted lowry acid
c) lewis base
d) lewis acid

A

b) bronsted lowry acid

147
Q

auto ionization of water is negligible if the conc of acid or base is….

A

significantly greater than 10-7 M

148
Q

auto ionization of water is important if the conc of acid or base is….

A

close to 10-7 M

151
Q

can pH be greater than 14 or less than 0

A

yes –> implies very strong conc

152
Q

weak acid in water eq

A

HA + H2O = H3O+ + A-

153
Q

the smaller the Ka, the _____ the acid, the _____ it will dissociate

A

weaker, less

154
Q

the smaller the Kb, the _____ the acid, the _____ it will dissociate

A

weaker, less

155
Q

weak base eq

A

BOH = B+ + OH-

156
Q

What are the pH and pOH of a solution containing 5 mL of 5 M benzoic acid (Ka = 6.3 x 10^-5) and 100 mL of 0.005 M benzoate solution?

A
157
Q

weak acid

Ka…

A

Ka < 1

158
Q

weak base

Kb…

A

Kb<1

159
Q

conjugate acid

A

acid formed when base gains proton

160
Q

conjugate base

A

base formed when acid loses proton

161
Q

removing a proton from a molecule produces…

A

conjugate base

162
Q

How does the common ion effect relate to Le Chatelier’s principle?

A

The common ion effect will shift the equilibrium by increasing the concentration of one of the products. This is an example Le Chatelier’s principle, where the equilibrium adjusts to relieve the added stress of increasing the products’ concentration.

163
Q

adding a proton from a molecule produces….

A

conjugate acid

164
Q

how are Ka and Kb related?

A

inversely related

165
Q

inert

A

unreactive (acid or base)

166
Q

acid strength

induction

A
  • electro neg elements near acidic proton increase acid strength
    • pull electron density out of bond holding the acidic proton
    • weakens proton bonding and facilitates dissociation
167
Q

when need to take square root…

A

adjust coefficient as needed to make power of 10 an even number

then sq rt only requires cutting the power of 10 in half

168
Q

hydrolysis

A

salt ions react with water to give back the acid or base

169
Q

a) 9
b) 6.1
c) 4
d) 7.1

A

a) 9

170
Q

a) IO4-
b) CN-
c) CH3CO2-
d) NO3-

A

b) CN-

171
Q

a) 1.2
b) 10.2
c) 9.1
d) 7.1

A

c) 9.1

172
Q

What is the pH of a buffer solution that contains 3.3 x 10^-5 M NH3 and 7.2 x 10^-5 NH4+ (Ka = 5.6 x 10^-10)?

a) 9
b) 7.6
c) 10.7
d) 8.3

A

a) 9

173
Q

a) 16
b) 0.06
c) 1.3 x 10^5
d) 7.9 x 10^-6

A

b) 0.06

174
Q

a) 12
b) 5
c) 2
d) 7

A

d) 7

175
Q

A student adds sodium acetate to a solution of acetic acid. What will happen to the pH and why?

a) the pH will decrease because an increased conc of acetate ions will push the acid dissociation reaction forwards
b) the pH will decrease because an increased conc of acetate ions will push the acid dissociation reaction backwards
c) the pH will increase because an increased conc of acetate ions will push the acid dissociation reaction forwards
d) the pH will increase because an increased conc of acetate ions will push the acid dissociation reaction backwards

A

d) the pH will increase because an increased conc of acetate ions will push the acid dissociation reaction backwards

176
Q

A researcher adds 0.25 mol of sodium acetate to 1 L of water. What will be the resulting pH? (Acetate has a Kb of 5.6 x 10^-10)

a) 9.1
b) 4.9
c) 12.5
d) 1.5

A

a) 9.1

183
Q

what the mathematical relationship between Ka, Kb, and Kw

A

Kw = Ka x Kb

185
Q

strong acids

A
  1. HI
  2. HCl
  3. HBr
  4. HNO3
  5. HClO3 (chloric acid)
  6. HClO4 (perchloric acid)
  7. H2SO4 (first proton only)
186
Q

strong bases

A
  1. Group I hydroxides
  2. Sr(OH)2
  3. Ca(OH)2
  4. Ba(OH)2
187
Q

Which of the following compounds can behave as a Bronsted-Lowry acid but not a Lewis acid?

(A) I2
(B) H2O
(C) HF
(D) None of the above

A

(C) HF

The answer must have a Hydrogen to donate, and cannot be able to accept any electron pairs. This is only represented by HF.

188
Q

Which is a stronger base, H2O or Cl-, in the following reaction: H2O + HCl H3O+ + Cl-

A

H2O is a stronger base than Cl- because Cl- is the conjugate base of a strong acid, making it a very weak base.

191
Q

What is pKw in terms of pH and pOH?

A

pKw (14) = pOH + pH

192
Q

Which of the following is the strongest base?

(A) F-
(B) CH3O-
(C) CH3COO-
(D) NO3-

A

(B) CH3O-

CH3OH is a very weak acid, so its conjugate base will be extremely strong.

193
Q

In terms of acids and bases, define a neutralization reaction.

A

A neutralization reaction is where an acid and a base react to form a salt. Oftentimes, water is also a product of these reactions.

207
Q

2 common buffers

A
  • acid buffer: acetic acid (CH3COOH) and its salt sodium acetate (CH3COO- Na+)
  • base buffer: ammonia (NH3) and its salt ammonium chloride (NH4+ Cl-)
208
Q

bicarbonate buffer system

A

H2CO3/HCO3-

  • carbonic acid (H2CO3) and its conjugate base form a weak acid buffer for maintaining the pH of the blood within a fairly narrow physiological range
    • usually caused by CO2 from breathing
209
Q

buffer range of optimal activity

A

pKa +/- 1

210
Q

when is buffering capacity optimal

A

half equivalence point

pH = pKa

211
Q

what happens when you change the ratio of the conjugate base to the acid in a buffer solution?

A

change in the pH of buffer solution

212
Q

what happens if the conc of both acid and conjugate base of buffer solution were double?

A

pH would not change

buffer capacity doubles -> addtn of small amount of acid or base will cause even less deviation in pH

213
Q

buffer capacity

A

ability to which the system can resist changes in pH

214
Q

buffering region of a titration curve

A
  • when [HA] ≈ [A-]
  • flattest portion of titration curve (resistant to changes in pH)
215
Q

what part of pH range will the equivalence points fall for weak acid + weak base?

A

equiv point can be in acidic, neutral, or basic range

depending on the relative strengths of acid and base

217
Q

solving buffer problem

A
  1. see what you’re adding reacts with in the buffer
  2. calculate moles
  3. make eq with part of buffer that is reacting with
    1. break down strong whatever into constituents first
  4. ICEBOX
  5. add concentrations of what adding and both parts of buffer
  6. solve
218
Q

What is the Kb of Cl- if the pKa of HCl is -7?

(A) 1 ⋅ 10^-21
(B) 1 ⋅ 10^-14
(C) 1 ⋅ 10^-7
(D) 1 ⋅ 10^7

A

(A) 1 ⋅ 10^-21

pKw = pKa + pKb
14 = -7 + pKb
21 = pKb
pKb = -log(Kb)
21 = -log(Kb)
Kb = 1 ⋅ 10^-21
219
Q

In the hydrolysis of salts, a cation and anion will be freed and may interact with water. Which of the following anions would be a stronger base than water?

(A) CN-
(B) Br-
(C) CH3CO2-
(D) NO3-

A

(A) CN-

The conjugate base to a strong acid will not react with water; the weaker that the acid is, the stronger its conjugate base is.

Both CH3CO2- and CN- are conjugate bases to weak acids (acetic acid and hydrogen cyanide, respectively). In this case, we must compare the acids’ pKa’s to water’s pKa: acetic acid has a pKa of 4.75, whereas hydrogen cyanide has a pKa of 9.2.

Water has a pKa of 7, so only CN- is a stronger base than water.

220
Q

Which of the following salts will NOT react with the water after hydrolysis?

(A) MgCl2
(B) Fe(NO3)3
(C) CuCl2
(D) NaNH2

A

(A) MgCl2

Copper (II) and Iron (III) are both stronger acids than water and will react with the water, and NH2- is a stronger base than water. On the other hand, Magnesium is a larger group 2 element and Cl- is the conjugate base to a strong acid, so neither the cation nor anion will react with the water!

221
Q

If the Ka of CH3COOH is 1.8 ⋅ 10^-5, what is the pH of a .54 M solution of CH3COO-?

(A) 10.35
(B) 9.24
(C) 8.43
(D) 7.21

A

(B) 9.24

Ka⋅Kb = Kw
(1.8 ⋅ 10^-5)Kb = 1 ⋅ 10^-14
Kb = (1 ⋅ 10^-14)/(1.8 ⋅ 10^-5)
Kb = approx. 5 ⋅ 10^-10 (actual: 5.6 ⋅ 10^-10)

Kb = ([CH3COOH][OH-])/[CH3COO-]
5 ⋅ 10^-10 = (x^2)/(.54-x) but we can just assume that x is negligible compared to .54:
5 ⋅ 10^-10 = (x^2)/(.54)
approx 2.5 ⋅ 10^-10 = x^2
x = [OH-] = approx. 1.35 ⋅ 10^-5 (actual: 1.74 ⋅ 10^-5)

pOH = -log([OH-])
pOH = -log(1.35 ⋅ 10^-5)
pOH = approx. 4.65 (actual: 4.76)
pKw = pOH + pH
14 = 4.65 + pH
pH = approx. 9.35 (actual: 9.24)
223
Q

What is the pH of a solution of 5.6 ⋅ 10^-1 M acetic acid (Ka = 1.8 ⋅ 10^-5) and 5.3 ⋅ 10^-2 M sodium acetate?

(A) 2.31
(B) 3.72
(C) 4.56
(D) 5.89

A

(B) 3.72

Ka = ([CH3COO-][H3O+])/[CH3COOH]
1.8 ⋅ 10^-5 = ([5.3 ⋅ 10^-2 + x][x])/[5.6 ⋅ 10^-1 - x] but x can be assumed to be negligible compared to our concentrations:
1.8 ⋅ 10^-5 = ([5.3 ⋅ 10^-2][x])/[5.6 ⋅ 10^-1]
1.8 ⋅ 10^-5 = approx. (1 ⋅ 10^-1)x
(1.8 ⋅ 10^-5)/(1 ⋅ 10^-1) = x
x = approx. 2 ⋅ 10^-4 (actual: 1.90 ⋅ 10^-4)

pH = -log([H3O+])
pH = -log(2 ⋅ 10^-4)
pH = 3.8 (actual: 3.72)
224
Q

Explain how a weak acid and its conjugate base make for a good buffer solution?

A

A weak acid will react with OH-, removing it from the solution. Its conjugate base will react with H+, removing it from the solution. This effectively protects our solution from an increase or decrease in the pH.

225
Q

But why not use a strong acid and its conjugate base?

A

The conjugate base of a strong acid is EXTREMELY weak and as such would not react with H+, leaving your solution vulnerable to decreases in pH.

226
Q

What is the pH of a solution that contains equal concentrations of its conjugate acid and its conjugate base (pKa = 5)?

(A) 4
(B) 5
(C) 6
(D) 7

A

(B) 5

The pH equals the pKa when the concentrations of acid and conjugate base are equal. You did not need the HH equation to solve this. :)

227
Q

You have a buffer solution containing a weak acid (pKa = 4.2) that has a pH of 7. A base is added, increasing the OH- concentration of the buffer solution by 10x the original amount. What is the new pH?

(A) -3
(B) 6
(C) 8
(D) 17

A

(C) 8

Because we are told that the hydroxyl concentration of the solution has increased tenfold, we do not need to worry about the buffering capabilities.

You can use the Henderson-Hasselbach equation to solve this type of problem, but you should be able to solve it conceptually just knowing that the pH scale is a logarithmic one.

228
Q

What is the pH of a buffer solution that contains 3.3 ⋅ 10^-5 M NH3 and 7.2 ⋅ 10^-5 M NH4+ (Ka = 5.6 ⋅ 10^-10)?

(A) 7.56
(B) 8.32
(C) 9.02
(D) 10.67

A

(C) 9.02

Don’t even waste your time pulling out the Henderson-Hasselbach equation for this one! The concentrations of the conjugate acid and base are less than a factor of ten apart and there is more acid than base, so the pH must be slightly less than the pKa (9.25). Most buffer solution questions on the MCAT should be solved intuitively without using the HH equation! :)

229
Q

solving weak acid/base + strong acid/base

A
  1. calculate moles
  2. determine how much reacts
    1. if = mol at end –> buffer –> other box (subtract and calculate M with new volume)
    2. if not = mol –> end product reacts again with H2O –> ICEBOX (solve for x)
230
Q

ammonium formulat

A

NH4+

231
Q

acetate formula

A

CH3COO-

232
Q

cyanide formula

A

CN-

233
Q

permanganate formula

A

MnO4-

234
Q

nitrate formula

A

NO3-

235
Q

nitrite formula

A

NO2-

236
Q

sulfate formula

A

SO42-

237
Q

sulfite formula

A

SO32-

238
Q

carbonate formula

A

CO32-

239
Q

phosphate formula

A

PO43-

240
Q

What is the pH of a 143.8 mL solution of .32 M NH3 (pKb = 4.75) when you add 45.9 mL of .13 M HCl?

(A) 5.61
(B) 6.98
(C) 8.32
(D) 10.43

A

(D) 10.43

  1. 9 mL HCl ⋅ 1 L / 1000 mL ⋅ .13 moles HCl / 1 L HCl = approx. 5 ⋅ 10^-3 moles HCl (actual: 5.967 ⋅ 10^-3)
  2. 8 mL NH3 ⋅ 1 L / 1000 mL ⋅ .32 moles NH3 / 1 L NH3 = approx. 45 ⋅ 10^-3 (actual: 46.016 ⋅ 10^-3)

45 ⋅ 10^-3 - 5 ⋅ 10^-3 = 40 ⋅ 10^-3 moles NH3

5 ⋅ 10^-3 moles NH4+

Once again, you can use the Henderson-Hasselbach equation here, but it will slow you down big time. Simply notice that the concentration of base (NH3) will be about 10x greater than the conjugate acid (NH4+), making the solution approximately 1 pH unit higher than the pKa of NH3.

Also notice that the pKb was given, not pKa (pKb = pKw - pKa = 14 - 4.75 = 9.25)

241
Q

The pH ranges for methyl red, bromthymol blue, and phenolphthalein are 4.4 - 6.2, 6.0 - 7.6, and 8.2 - 10, respectively. Which would best be used for titrating a strong acid with a strong base in which the steep part of the titration curve ranges from a pH of 1.8 to 12.7?

I. Methyl red
II. Bromthymol blue
III. Phenolphthalein

(A) I Only
(B) II Only
(C) III Only
(D) I, II, and III

A

(D) I, II, and III

Because the curve is so steep, any one of these indicators would change color near the equivalence point.

242
Q

True or false? For a polyvalent acid or base, there will be one elongated titration curve and a single endpoint.

A

False. For a polyvalent acid or base, there will be an endpoint for each of the acid or base equivalents, and typically some separation between one endpoint and the next titration.

243
Q

What is the purpose of a redox titration vs. an acid/base titration?

A

THEY HAVE THE SAME PURPOSE! ;) The purpose is to find the concentration of a solution. The difference between these methods is not their purpose but their methodology. One uses the equivalence point between an acid and its base while the other uses the equivalence point between an oxidized species and its reduced species.

244
Q

Consider the following unbalanced reaction that results from titrating a 24.3 mL solution of Fe2+ with 67.9 mL of a 5.63 ⋅ 10^-3 M solution of KMnO4: MnO4- + Fe2+ -> Mn2+ + Fe3+ What is the concentration of the original Fe2+ solution?

(A) .0786 M
(B) .0324 M
(C) .9425 M
(D) .3652 M

A

(A) .0786 M

First, balance the reaction as follows: MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O

  1. 9 mL MnO4- ⋅ 1 L / 1000 mL ⋅ 5.63 ⋅ 10^-3 moles MnO4- / 1 L MnO4- = approx. 385 ⋅ 10^-6 moles MnO4-
    (actual: 382.277 ⋅ 10^-6)

385 ⋅ 10^-6 moles MnO4- ⋅ 5 moles Fe2+ / 1 mole MnO4- = approx. 2000 ⋅ 10^-6 moles Fe2+ (actual: 1911 ⋅ 10^-6)

2 ⋅ 10^-3 moles Fe2+ / 2.43 ⋅ 10^-2 L Fe2+ = approx. 1 ⋅ 10^-1 M (actual: .786 ⋅ 10^-1)

245
Q

chlorate formula

A

ClO3-

246
Q

chlorite formula

A

ClO2-

247
Q

hydrogen carbonate (bicarbonate) formula

A

HCO3-

248
Q

hydrogen phosphate formula

A

HPO4(2-)

249
Q

chromate formula

A

CrO4(2-)

250
Q

Write the net ionic equation for adding a solution of NaCl to a solution of AgNO3 to form the precipitate AgCl.

A

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

251
Q

hypochlorite formula

A

ClO-

252
Q

thiocynate

A

SCN-

253
Q

dichromate formula

A

Cr2O7(2-)

254
Q

oxalate formula

A

C2O4(2-)

255
Q

thiosulfate formula

A

S2O3(2-)

256
Q

peroxide formula

A

O2(2-)

257
Q

a) the less acidic hydrogen, bc its pka = 10.33, allowing carbonic acid to act as a buffer at blood’s slightly acidic pH
b) the less acidic hydrogen, bc its pka = 10.33, allowing carbonic acid to act as a buffer at blood’s slightly basic pH
c) the more acidic hydrogen, bc its pka = 6.35, allowing carbonic acid to act as a buffer at blood’s slightly basic pH
d) the more acidic hydrogen, bc its pka = 6.35, allowing carbonic acid to act as a buffer at blood’s slightly acidic pH

A

c) the more acidic hydrogen, bc its pka = 6.35, allowing carbonic acid to act as a buffer at blood’s slightly basic pH

258
Q

sodium carbonate is commonly used to soften water. what is the ionic formula of sodium carbonate?

a) Na2CO3
b) NaCO2
c) NaCO3
d) Na2CO2

A

a) Na2CO3

259
Q

calcium chlorate can be used as part of weedkillers. what is the ionic formula for this salt?

a) Ca(ClO4)2
b) CaClO3
c) CaClO4
d) Ca(ClO3)2

A

d) Ca(ClO3)2

260
Q

what processes will occur when solid NaCl and AgNO3 are added to water?

I. Precipitation

II. Dissolution

III. Hydration

a) I, II, and III
b) I only
c) II and III only
d) I and II only

A

a) I, II, and III

261
Q

a) neither spectator ions nor catalysts affect the equilibrium of the rxn
b) spectator ions are not usually involved in the rxn, whereas catalysts are
c) both spectator ions and catalysts are left out of net ionic equations
d) both spectator ions and catalysts affect the rate of reaction

A

d) both spectator ions and catalysts affect the rate of reaction

262
Q

a) (NH4)HPO4-
b) (NH4)2HPO4
c) (NH4)2H2PO4
d) (NH4)H2PO4

A

b) (NH4)2HPO4

263
Q

NH3 has a pKb of 4.75. Would methyl red or phenolphthalein be a better indicator for titrating NH3 with HCl and why?

a) Phenolphthalein, because the half equivalence point overlaps with methyl red’s indicator range of pH 4-6.
b) Methyl red, because the equivalence point overlaps with phenolphthalein’s indicator range of pH 8-10.
c) Methyl red, because the half equivalence point overlaps with phenolphthalein’s indicator range of pH 8-10.
d) Phenolphthalein, because the equivalence point overlaps with methyl red’s indicator range of pH 4-6

A

c) Methyl red, because the half equivalence point overlaps with phenolphthalein’s indicator range of pH 8-10.

264
Q

a) I and III only
b) II only
c) III only
d) I, II, and III

A

d) I, II, and III

265
Q

a) 5.3
b) methyl red cannot be used to indicate equivalence for acids
c) 8.4
d) 2.8

A

d) 2.8

266
Q

a) 3.85
b) 8.35
c) 10.15
d) 5.65

A

c) 10.15

267
Q

using the HH eq, how does the pH change based on the acid/base conc

A

the pH increases by 1 from its pKa for each factor of 10 by which the base’s conc is greater than the acid’s conc

MCAT (26 decks)

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