Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

MCAT > Chemistry: Acids & Bases > Flashcards

Chemistry: Acids & Bases Flashcards

You may prefer our related Brainscape-certified flashcards:
MCAT flashcards
1
Q

Strong Acids

A

Strongest: HCl, H2SO4, HNO

HBr, HI, HClO4 (HPO4?), HClO3

Strong acid	Formula
Perchloric acid	HClO4
Hydroiodic acid	HI
Hydrobromic acid	HBr
Sulfuric acid	H2SO4
Hydrochloric acid	HCl
Nitric acid	HNO3
Hydronium ion	H3O+ or H+
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Strong Bases

A 
Strong bases	Formula
Lithium hydroxide	LiOH
Sodium hydroxide	NaOH
Potassium hydroxide	KOH
Rubidium hydroxide	RbOH
Cesium hydroxide	CsOH
Calcium Hydroxide	Ca(OH)2
Strontium hydroxide	Sr(OH)2
Barium hydroxide	Ba(OH)2
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is the ionization constant of H2S?

A

much less than 1

WEAK ACID!

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

-SH vs. -OH

A

Because of charge density (charge per volume):

  • SH: weak
  • OH: strong

Share ~same charge but S has larger volume it has lower charge density… means more stable and therefore less reactive
charge density= charge/volume

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

pH

A

-log[H+]

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Assuming 25 degrees Celsius:

pH + pOH =

A

14

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q 
What is the relationship between the variables on the L and R sides of the equations?
pH=-log[H+]
pOH=-log[OH]
pKa = -log[Ka]
pKb = -log[Kb]
A

inverse:
ie: if pH goes down, [H+] goes up

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

a) What is pH if [H+] = 1.0 x 10^-1?
b) What is pOH if 1.0 x 10^8

(@25 degrees C)

A

a) 1

b) -8

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Steps to convert concentrations of H+ and OH- into pH and pOH (@25 degrees C)

Examples
A) H+: 6.4 x 10^-3
B) OH-: 2.1 x 10^-8

A

A) H+: 6.4 x 10^-3
1) For ones value of answer (pH from [H+]): Lock in on the power of 10. Flip the sign and subtract 1.
(10^-3) –> -3 –> +2

2) For decimal of answer: Subtract the digits (before the power of 10) from 10.
(6. 4 ~ 6) –> 10-6 = 4

3) Put it together: 2.4 = pH
_______________
B) OH-: 2.1 x 10^-8
pOH=7.8

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

(In 3 steps)
Convert pH into [H+]
a) pH: 13.3
b) pH: 1.8

(also works for pOH–>OH, pKa–>Ka, pKb–Kb, @25 degrees C)

A

A)pH: 13.3

 1) 13 --> 10^-14
 2) .3 -->10-3 = 7
 3) 7 x 10^-14 = [H+]

A)pH: 1.8

 1) 1 --> 10^-2
 2) .8 -->10-8 = 2
 3) 2 x 10^-2 = [H+]
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q 
Based on the pKa, which of the following species is the strongest base?
pKa of ammonium ion for:
*NH3:  9.26
*(CH3CH2)2NH: 10.64
*(Ch3CH2)2NH: 10.98
*(CH3CH2)3N: 10.76
A

Answer: (CH3CH2)2NH:

pKa –> pKb of ammonium ion: (14-pKa=pKb)

So,

  • NH3: 9.26 (pKb: 4.74)
  • (CH3CH2)2NH: 10.64 (pKb: 3.36)
  • (CH3CH2)2NH: 10.98 (pKb: 3.02)
  • (CH3CH2)3N: 10.76 (pKb) 3.24

Smallest Kb: 4.74
*Largest Kb: 3.02

Largest pKa = Smallest pKb = Largest Kb = strongest base

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

The bigger the pKa, the ____ the base?

A

Bigger the pKa –>
Smaller the Ka

Smaller the Ka –>
Bigger the Kb

Bigger the Kb –>
Stronger the base

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Reacting equivalents

A

make stoich corrections for how many mol acid/base are produced per mol of reacting source

Ex:
HCl –> H+ + Cl-
(1 mol HCl –> 1 eq acid)

H2SO4 –> 2H+ + SO4^2-
(1 mol HsSO4 –> 2 eq acid)

Ca(OH)2 –> Ca^2+ + 2OH-
(1mol Ca(OH)2 –> 2 eq base)
———————-
(MCAT limits to Acid-Bases… but it applies elsewhere)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Gram-Equivalent Weight

Ex: NaOH and Na2CO3

A

Stoich correction of molecular weight
g/eq = g/mol X mol/eq

NaOH (1mol:1eq ratio)
40 g/mol x 1 mol/1eq= 40 g/eq

Na2CO3 (1mol:2eq ratio)
106 g/mol x 1 mol/2 eq = 53 g/eq

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Normality (N)

Ex: H2SO4

A

stoich corrections for molality–needed for world of equivalents in world of eq points (as opposed to world of mols)

eq/L = mol/L x eq/mol

1O^-1 mol/L x 2 eq/mol = 2 x 10^-1 eq/L

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Titration Curves

A

Strong Acid-Strong Base: steeper

Weak Acid-Strong Base: less steep

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Titrant

A

solution of known concentration

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Equivalence Point Formula

A

WHEN you are at the equivalence point:
Eq acid = Eq base

  • Use this formular to find the M? or vol? or grams?
  • Does not care about pH if asking for volume of base added!!!
  • If everything is monoprotic/monobasic, you can interchange M and N… but be read JIC
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Eq Pt Formula: Both aq

A

(Normality of Acids)(Volume of Acid) = (Normality of Base)(Volume of Base)
(eq/L)(L) = (eq/L)(L)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Eq Pt Formula: aq/solid

A

(Normality of Acids)(Volume of Acid) = mass base/gram eq wt = mass/mw x eq/mol

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Titration Question Types

pH question families

A

1) What is the pH at X point in the curve?

2) What is the volume/grams/moles of the solution ?

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Eq Pt Formula: if using both solids

A

mass of acid/gram eq wt = mass of base/ gram eq wt

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Set up the formula to find:

How many mg of NaOH (s) is required to neutralize 408 mg of KHP (s), a monoprotic weak acid with a mw of 204 amu?

A

Monoprotic means Gram Eq Wt (GEW)= mw

Mass of acid/ GEW = mass of base / GEW

408/204 = x/40
2=x/40
x=80

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

Set up the formula to find:

How many mg of NaOH (s) is required to neutralize 40 mL of a 10^-1 M HCl solution?

A

(N of acid) * (V of acid) = mass/GEW

(10^-1)(40) = x/40
x=160

25
Q

In Weak Acid-Strong Base (or SA-WB–same)Titrations, how are the following affected?

  • pH initial
  • pH Halfway
  • pH at eq pt

e.g.
HA + NaOH –> H2O + A-Cl+

A

Buffer solution: weak acid with its weak CB in solution at the same time (Henderson-Hasselbach time)

Simulation:

  • Initial: HCl=30eq, NaCl=0eq
  • Halfway: HCl=15eq, NaCl=15eq
  • Eq Pt: HCl=0eq, A-Cl+=30eq
  • pH initial: must use Ka expression to solve
  • pH Halfway: buffer soln (must use H-H eqn to solve)
  • pH at eq pt: HA neutralized, A- = wk base, pH > 7
25
Q

In Weak Acid-Strong Base (or SA-WB–same)Titrations, how are the following affected?

  • pH initial
  • pH Halfway
  • pH at eq pt

e.g.
HA + NaOH –> H2O + A-Cl+

A

Buffer solution: weak acid with its weak CB in solution at the same time (Henderson-Hasselbach time)

Simulation:

  • Initial: HCl=30eq, NaCl=0eq
  • Halfway: HCl=15eq, NaCl=15eq
  • Eq Pt: HCl=0eq, A-Cl+=30eq
  • pH initial: must use Ka expression to solve
  • pH Halfway: buffer soln (must use H-H eqn to solve)
  • pH at eq pt: HA neutralized, A- = wk base, pH > 7
26
Q

In Strong Acid-Strong Base Titrations, how are the following affected?

  • pH initial
  • pH Halfway
  • pH at eq pt

e.g.
HCl + NaOH –> H2O + NaCl

A

*Assume 100% disassociation.

Simulation:

  • Initial: HCl=30eq, NaCl=0eq
  • Halfway: HCl=15eq, NaCl=15eq
  • Eq Pt: HCl=0eq, NaCl=30eq
  • pH initial: easy to solve bc [HCl]initial = [H+]
  • pH Halfway: NaCl is inert, so [HCl]halfway = [H+]
  • pH at eq pt (look at the products): HCl neutralized, NaCl inert, pH = 7

Aside: how do we know NaCl is inert? it is a conjugate of strong species. That is HCl is strong so produces inert Cl. NaOH is strong so produces inert Na.

26
Q

In Strong Acid-Strong Base Titrations, how are the following affected?

  • pH initial
  • pH Halfway
  • pH at eq pt

e.g.
HCl + NaOH –> H2O + NaCl

A

*Assume 100% disassociation.

Simulation:

  • Initial: HCl=30eq, NaCl=0eq
  • Halfway: HCl=15eq, NaCl=15eq
  • Eq Pt: HCl=0eq, NaCl=30eq
  • pH initial: easy to solve bc [HCl]initial = [H+]
  • pH Halfway: NaCl is inert, so [HCl]halfway = [H+]
  • pH at eq pt (look at the products): HCl neutralized, NaCl inert, pH = 7

Aside: how do we know NaCl is inert? it is a conjugate of strong species. That is HCl is strong so produces inert Cl. NaOH is strong so produces inert Na.

27
Q

What is the pH of a 10^-3 HCl soln (pKa = -8)

A

Strong acid soln: 100% disassociation

[HCl]initial –> [H3O+]
[H+] = 10^-3
pH = 3

27
Q

What is the pH of a 10^-3 HCl soln (pKa = -8)

A

Strong acid soln: 100% disassociation

[HCl]initial –> [H3O+]
[H+] = 10^-3
pH = 3

28
Q

What is the pH of a 10^-3 M CH3COOH soln (pKa = 5)?

A

-less likely to be on MCAT

Ka=[A-][H3O+]/[HA]

HA + H3O –> A- + H3O+
-x x x
__________________

10^-5 = x^2/10^-3
x^2 = 10^-8
x=10^-4 = H+
pH = 4
28
Q

What is the pH of a 10^-3 M CH3COOH soln (pKa = 5)?

A

-less likely to be on MCAT

Ka=[A-][H3O+]/[HA]

HA + H3O –> A- + H3O+
-x x x
__________________

10^-5 = x^2/10^-3
x^2 = 10^-8
x=10^-4 = H+
pH = 4
29
Q

What is the pH of a 10^-3 M NH3 soln (pKb = 5)

A

-less likely to be on MCAT bc of math

Kb=[HB+][OH-]/[B]
10^-5 = x^2/10^-3
x=10^-4 = Oh-
pOH = 4
pH = 10
29
Q

What is the pH of a 10^-3 M NH3 soln (pKb = 5)

A

-less likely to be on MCAT bc of math

Kb=[HB+][OH-]/[B]
10^-5 = x^2/10^-3
x=10^-4 = Oh-
pOH = 4
pH = 10
30
Q

What is the pH of a 10^-3 M HCl at the 1/2 eq pt when titrated with NaOH (s)?

A

-SA 1/2 eq pt less likely to be asked on MCAT bc pH barely budges from initial

[H+]=(10^-3)/2 = 5 x 10^-4
pH=3.5

30
Q

What is the pH of a 10^-3 M HCl at the 1/2 eq pt when titrated with NaOH (s)?

A

-SA 1/2 eq pt less likely to be asked on MCAT bc pH barely budges from initial

[H+]=(10^-3)/2 = 5 x 10^-4
pH=3.5

31
Q

pH at eq pt

A

MCAT will ask in terms of acid and basic.

31
Q

pH at eq pt

A

MCAT will ask in terms of acid and basic.

32
Q

Ammonia Buffer Solution

A

Finish! slide 18

HH in head

32
Q

Ammonia Buffer Solution

A

Finish! slide 18

34
Q

What is the pH of a 10^-3 M CH3COOH (pKa=5) at the 1/2 eq pt when titrated with NaOH (s)?

A

pH=5

  • Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln
  • Buffer soln –> use HH eqn!

pH = pKa + log (A-/HA) –> at 1/2 eq pt, pH = pKa

pH=5

36
Q

What is the pH of a 10^-3 M CH3COOH (pKa=5) at the 1/2 eq pt when titrated with NaOH (s)?

A

pH=5

  • Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln
  • Buffer soln –> use HH eqn!

pH = pKa + log (A-/HA) –> at 1/2 eq pt, pH = pKa

pH=5

38
Q

What is the pH of a 10^-3 M NH3 (pKb=5) at the 1/2 eq pt when titrated with NaOH (s)?

A

pH=9

-Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln

pOH = pKb + log (HB+/B) –> at 1/2 eq pt, pOH = pKb

pOH=5,
pH = 14-5 = 9

40
Q

What is the pH of a 10^-3 M NH3 (pKb=5) at the 1/2 eq pt when titrated with NaOH (s)?

A

pH=9

-Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln

pOH = pKb + log (HB+/B) –> at 1/2 eq pt, pOH = pKb

pOH=5,
pH = 14-5 = 9

45
Q

In Weak Acid-Strong Base (or SA-WB–same)Titrations, how are the following affected?

  • pH initial
  • pH Halfway
  • pH at eq pt

e.g.
HA + NaOH –> H2O + A-Cl+

A

Buffer solution: weak acid with its weak CB in solution at the same time (Henderson-Hasselbach time)

Simulation:

  • Initial: HCl=30eq, NaCl=0eq
  • Halfway: HCl=15eq, NaCl=15eq
  • Eq Pt: HCl=0eq, A-Cl+=30eq
  • pH initial: must use Ka expression to solve
  • pH Halfway: buffer soln (must use H-H eqn to solve)
  • pH at eq pt: HA neutralized, A- = wk base, pH > 7
46
Q

In Strong Acid-Strong Base Titrations, how are the following affected?

  • pH initial
  • pH Halfway
  • pH at eq pt

e.g.
HCl + NaOH –> H2O + NaCl

A

*Assume 100% disassociation.

Simulation:

  • Initial: HCl=30eq, NaCl=0eq
  • Halfway: HCl=15eq, NaCl=15eq
  • Eq Pt: HCl=0eq, NaCl=30eq
  • pH initial: easy to solve bc [HCl]initial = [H+]
  • pH Halfway: NaCl is inert, so [HCl]halfway = [H+]
  • pH at eq pt (look at the products): HCl neutralized, NaCl inert, pH = 7

Aside: how do we know NaCl is inert? it is a conjugate of strong species. That is HCl is strong so produces inert Cl. NaOH is strong so produces inert Na.

47
Q

What is the pH of a 10^-3 HCl soln (pKa = -8)?

A

Strong acid soln: 100% disassociation

[HCl]initial –> [H3O+]
[H+] = 10^-3
pH = 3

48
Q

What is the pH of a 10^-3 M CH3COOH soln (pKa = 5)?

A

-less likely to be on MCAT

Ka=[A-][H3O+]/[HA]

HA + H3O –> A- + H3O+
-x x x
__________________

10^-5 = x^2/10^-3
x^2 = 10^-8
x=10^-4 = H+
pH = 4
49
Q

What is the pH of a 10^-3 M NH3 soln (pKb = 5)

A

-less likely to be on MCAT bc of math

Kb=[HB+][OH-]/[B]
10^-5 = x^2/10^-3
x=10^-4 = Oh-
pOH = 4
pH = 10
50
Q

What is the pH of a 10^-3 M HCl at the 1/2 eq pt when titrated with NaOH (s)?

A

-SA 1/2 eq pt less likely to be asked on MCAT bc pH barely budges from initial

[H+]=(10^-3)/2 = 5 x 10^-4
pH=3.5

51
Q

What is the pH of a 10^-3 M CH3COOH (pKa=5) at the 1/2 eq pt when titrated with NaOH (s)?

A

pH=5

  • Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln
  • Buffer soln –> use HH eqn!

pH = pKa + log (A-/HA) –> at 1/2 eq pt, pH = pKa

pH=5

52
Q

What is the pH of a 10^-3 M NH3 (pKb=5) at the 1/2 eq pt when titrated with NaOH (s)?

A

pH=9

-Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln

pOH = pKb + log (HB+/B) –> at 1/2 eq pt, pOH = pKb

pOH=5,
pH = 14-5 = 9

53
Q

pH at eq pt

A

MCAT will ask in terms of acid and basic.

54
Q

Ammonia Buffer Solution

A

Finish! slide 18

HH in head

55
Q

pH exercises

slide 20

A 
1) Ask: is it a buffer or not?
   Yes? HH time
2) No? Is acid or base winning?
    And by how much?
  If acid is winning by a factor of 10, for example, pH lowers by 1 pH point
56
Q

Buffer Capacity

A

1) look at eqs of Acid and CB
* Ability to resist change has everything to do of absolute amt equivalents of acid and CB (not identity of buffer acids/bases or pH value) –> more you have of both–> more capable you are of resisting pH changes due to outside influx of acid/base

if 1:1 ratio, pH = pKa

57
Q

Buffer Capacity:
How will the following solutions change pH after adding 10 eq H+?

CH3COO- + H3O+ –> CH3COOH + H2O

A

insert jpeg

58
Q

Indicators:

pKa of methyl red is 5.2
Describe the relative concentration of the colored form of a 1M solution of methyl red at pH 4.2, 5.2, 6.2

(insert jpeg if desire)

A

pH = pKa + log(yellow/red)

  1. 2 = 5.2 + log (1/10) RED
  2. 2 = 5.2 + log (1/1) ORANGE
  3. 2 = 5.2 + log (10/1) YELLOW

[Takeaway: color change occurs when pH solution is around pKa of an indicator, thus, choose an indicator whose pKa is NEAR ESTIMATED eq pt pH]

59
Q

A student was given a flask containing 25.0 mL of an aq soln of ammonia (pKb at 25 deg 4.74) of unknown concentration. The student was asked to determine the concentration of this solution by titrating it against a standard 0.10 M aq HCl soln that was placed in burette. The student added 2 drops of indicator to the soln and titrated it against the HCl solution. The experimental temp was 25 deg C. The student noted that 40.0 mL of the HCl solution was needed to reach the eq pt. Table 1 lists that pH range of commonly used acid-base indicators.

A

insert jpeg

pOH = pKb,
pH=pKa at top of curve 1/2 eq pt

MCAT (24 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions