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Bio/BC > 5A - Acid-Base Equilibria > Flashcards

5A - Acid-Base Equilibria Flashcards

1
Q

Mnemonic: Brotons-Lowry and Lewis model

A

Therefore Protons: Bronsted-Lowry

And remember Lewis dot model is drawing of electrons

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2
Q

What is an amphoteric species?

A

This is a species in which it has the ability to function both as an acid and base/ proton donor and proton acceptor/ electron acceptor and electron donor respectively.

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3
Q

Find the molarity of OH- if the concentration of H3O+ is 2.2x10-3 M.

A

H2O + H2O ⇔ H3O+ + OH-
Kw = KH3O+KOH- = 1x10-14 M / 2.2x10-3
~0.5 x 10-11 => 5 x 10-2M

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4
Q

Why is the understanding of water so important on the MCAT?

A

It is important to understand the behavior of acidic and basic compounds in water and because most reactions function in water.
This is why we need to understand the behavior of water and it pH

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5
Q 
Which of the following thermodynamic quantities is important in the value of Kw?
A. Concentration
B. Pressure 
C. Volume
D. Temperature
A

D. Temperature. @ 25ºC or 298K, the Kw = 10-14 Increase in temperature from 298, leads to an increase in Kw value and decrease in temperature from 298K leads to a decrease in Kw value
While the temperature of 298K does not change, the [ ] of H3O+ and OH- does have the ability to change, but Kw will never change
Others do not affect Kw

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6
Q

Lemon juice to have [H3O+] = 2.2 x 10-3 M, what is the OH- [ ]? Is this an acidic, basic, or neutral solution?

A

[H3O+] [OH-] = 1 x 10-14 M
1 x 10-14 M = 2.2 x 10-3 M [OH-]
[OH-] = 1 x 10-14 M / 2.2 x 10-3 M = 4.45 x 10-12 M
[H3O+] vs [OH-]
2.2 x 10-3 M vs 4.45 x 10-12 M
Because 2.2 x 10-3 M > 4.45 x 10-12 M, the solution is in fact acidic.

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7
Q

Explain the factors that can change Kw.

A

The magnitude of Kw like other equilibrium constants, will not unless the temperature changes. This means that the magnitudes of OH- and H3O+ will while the solution is neutral.
Increase in changes in temperature leads to an increase in Kw in value [likely due to the endothermic nature of autoionization.
Changes to the Kw value, changes the significance of the pH scale. This is why it is important in looking at the conditions presented to you on the test

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8
Q

Identify which reactant are amphoteric species in the following reactions. For the species, determine if the compound is amphoteric: HCO3- + HBr ⇔ H2CO3 + Br

A

Amphoteric reactant: HCO3-

This species is amphiprotic as well

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9
Q

Identify which reactant are amphoteric species in the following reactions. For the species, determine if the compound is amphoteric: 3HCl + Al(OH)3 ⇔ AlCl3

A

Amphoteric species: Al(OH)3 Note: HCL is not able accept as a base, but it does function as an acid. Note: this definition is not in reference to its conjugate species
This is not an amphiprotic species because it does not release H+ into solution

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10
Q

Identify which reactant are amphoteric species in the following reactions. For the species, determine if the compound is amphoteric: 2HBr + ZnO ⇔ ZnBr2 + H2O

A

The amphoteric species is ZnO

This is not an amphiprotic species because it does not release H+ into solution

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11
Q

What are the missing associated variables with a solution with a pH of 4: [H3O+], pOH. [OH-], acid or base?

A 
pH = 4
[H3O+] = 1 x 10-4 M
pOH = 10 
[OH-] = 1x10-10M
This is an acid! This is determined by the pH
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12
Q

A compound with 8.89x10-4M H3O+. What are the associated: pH, pOH, and [OH-]? Is this an acid or base?

A 
[H3O+] = 8.89x10-4
pH = -log(8.89x10-4) = ~4-0.889 = ~3.2 
pOH = 14 - 3.2 = ~10.8
[OH-] = 1x10-14 / 8.89x10-4  = ~0.112 x 10-14- -4 = 0.112 x 10-10 = 1.12 x 10-9 M 
This compound is an acid!
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13
Q

You’re working with an unknown element in lab and find their pOH = 5.19. Find the associated pH, H+ and OH- concentrations. Is this compound an acid or base?

A 
pOH = 5.19 
pH = 14 - 5.19 = ~ 8.8
[H3O+] = Remember that pH = -log[H+] = 8.8 => -1x10-8.8 M ~ 1x10-9 M (rounding the fractional exponents can allow you to give a good approximation)
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14
Q 
What is the pH when you have an [OH-] concentration equal to 9.84x10-8 M? 
A. 8.67 
B. 7.92
C. 6.99
D. 5.43
A

pOH = -log[OH-] = -log(9.84x10-8)
8 - 0.984 = 7.016
14 - 7.016 = 6.98 = pH AKA C. 6.99

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15
Q

If orange juice has a hydronium concentration of 3.2x10-4M, what is the pH?

A

pH = -log[3.2x10-4]
~ 4 - 0.32 = 3.68
= 3.5

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16
Q

Calculate the pH of an aqueous solution that contains 0.11g of Ca(OH)2 in a total volume of 250 mL
Ca(OH)2 -> Ca2+ + 2OH- | Note: This metal ion completely dissociates, this means that the total concentration of OH made in the product is double

A 
Molar mass of Ca(OH)2 = 74g/mol
0.11gCa(OH)2 / (74g/mol) = 0.0014 mol
0.0014 mol/250mL = 0.0056 M CaOH2
[OH-] = 0.0056 M CaOH2 (2OH-/CaOH2) = 0.0112 M OH
pOH = -log[OH-] = -log[0.0112] = 1.95
pH + pOH = 14
14 - 1.95 = 12.05 = pH

The Bicarb system favors the reactants more and makes for a very good buffer system hence why it is found in the blood to combat CO2 levels.

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17
Q

Predict is the conjugate base of HCl will be a strong or weak?

A

Remember that the stronger the acid/base, the weaker the conjugate base/acid.
think of this as competing base ranks Cl will try to pick up H on the product side. However since HCl is so good at donating an electron, the Cl will have a hard time doing what it desires. This is why strong acids and bases dissociate completely in one direction only and there is no reverse reaction because the conjugate acid is weak

18
Q

The pH of 0.03M solution of nitric acid in water. What is the pH of the solution?

A

HNO3 (nitric acid) + H2O -> NO3- + H3O+
Remember that this is a strong acid and will completely dissociate. This is demonstrated by a single headed arrow
This also means that there are 0.03M of NO3- and 0.03M of H3O+ produced
pH = -log[H3O+] = -log[0.03] = -log[3x10-2] = ~2-0.3 = 1.7

19
Q

NaCH3COO is added into neutral pH of water. What are the results?

A

when the salt dissociates into Na and CH3COO: Na will not affect the pH
CH3COO- - this will change pH. Why? Acetic acid is a weak acid, and when weak acids are dissociated, they make strong conjugate bases! This means that the ion formed form the salt will strip the H from the H2O, creating lots of OH- readily, producing a basic solution
NaCH3COO + H2O -> Na+ + OH- CH3COOH
Remember that Na and OH are conjugates of a strong acids, and therefore are weak and will not tend to interact with one another
Increase in OH- leads to an increase in pH
This means that a salt formed from a WEAK acid and a STRONG base, you’ll produce a basic solution

20
Q

This salt ClNH4 is added into neutral pH water

A

Remember Cl- does not change the pH of the solution much
NH4+ will affect the pH of the water by a lot. Weak base NH3 creates a strong conjugate acid NH4+ This means that in water, this acid will donate the H+ to water, to form H3O+
ClNH4 + H3O+ -> NH3 + Cl- + H3O+
Therefore when the salt of a STRONG acid and a weak base is dissolved in water, it will produce an acidic solution

21
Q

Calculate the concentration of H3O+ in a 2.0M aqueous solution of acetic acid, CH3COOH. (Note: Ka1.8x10-5)

A 
CH3COOH(aq) ⇔ CH3COO-(aq) + H3O+ (aq) 
2.0M			0M		0M
	-x			+x		+x
	2.0 - x		+x		+x
pKa = [H3O+][CH3COO-]/[CH3COOH]
pKa = x2/2.0M-x
Assume x far smaller than 2.0
pKa = x2 / 2
x2 = pKa*2 => 1.8x10-5*2 
x = √(3.6x10-6*) = 6x10-3 H3O+
22
Q

If a compound has a Ka value&raquo_space; Water, what does it mean about its behavior in solution? How does this compare with a solution that has only a slightly higher Ka than water?

A

Compound with a Ka&raquo_space; water means that it will behave as a strong acid, therefore it will dissociate completely in water.
Having a Ka slightly greater than water means the acid is a weak acid with minimal dissociation

23
Q

If a compound has a Kb value&raquo_space; water, what does it mean about its behavior in solution? How does this compare with a solution that has only a slightly higher Kb than water?

A

This compound with a very high Kb value compared to water means that this base is a strong base and therefore will dissociate completely in water.
Having a Kb slightly greater than water means that this compound is a base, but a very weak base and therefore will not dissociate completely but only partially.

24
Q

What is the acid dissociation constant for this equation: H2O + HCl => H3O+ + Cl-

A

In this example the BLB - Water - this is the molecule that accepts the H
BLA - HCl, there this is the molecule that donates the H
Strong acids - donate H very easily. This means that 100% of the reactants will become products
Ka = [H3O+] [Cl-] / [HCl]
The dissociation value will be based on how much product is made from the reaction. Because strong acids dissociate completely, Ka will be very large for forward reactions
Large value in numerator / small number in the denominator
In conclusion Ka > > 1 for strong acids

25
Q

What is the dissociation constant for this reaction: Acetic Acid + H2O => Acetate + H3O+

A

BLA - Acetic acid
BLB - water
Weak acids - acetic acid is a weak acid. Weak acids tend to not donate H+ well therefore they will produce a much stronger conjugate base. This creates a reaction in which acetic acid will stay mostly protonated, and therefore reactant favored
At equilibrium, there is a high concentration of the reactant to the products
Ka = dissociation
= [Products] / [reactants] = [small value]/ [large value]
= [CH3COO-] [H3O+] / [CH3COOH]
Ka &laquo_space;1 in conclusion for weak acids

26
Q

What is the expression for the equilibrium constant for this reaction: 3H2(g) + N2(g) ⇔ 2NH3(g)

A

Keq=[NH3]2 / ([H2]3[N2]
Because this reactions involves the gases, these will no be in terms of concentration, but instead in partial pressure!
Kp = PNH32 / PH23 PN2

27
Q

What is the pH of a buffer solution that is 0.24 M NH3 and 0.20 M NH4Cl?

A

NH3 + H2O ⇔ NH4+ + OH-
I 0.24 0.2 0
C -x +x +x
E 0.24-x 0.2+x +x

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

Kb = [NH4+][OH-]/[NH3] 
1.8x10-5 = (0.2+x)(x)/(0.24-x)
Assume x<<0.2
1.8x10-5 = 0.2x/0.24
x = 2.16x10-4 M OH-
pOH = -log[OH-] = -log[2.16x10-4] = 4 - 0.216 = 3.8 
pH + pOH = 14
14 - 3.8 = 10.2 = pH 
Note: this equation does not describe the pH of the buffer system. This method of solving is to describe how the pH changes when you add in a salt into an already existing solution. For working with buffer systems - use the henderson hasselbalch equation for buffer systems. The Henderson-Hasselbalch equation is useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid-base reaction. This means that this equation is used to find the pH at EQUILIBRIUM

In our equation, the acid/HA is NH4 - the conjugate acid and the base/A- is NH3

Kb = 1.8x10-5 
Ka*Kb = Kw 
Kw/Kb = Ka = 10-14/1.8x10-5 = 5x10-10 M 
pH = pKa + log[A-]/[HA]
= -log(5x10-10) + log(0.24/0.2) 
(10 - 0.5) - 0.92
= 9.5 - 0.079
pH = ~9.42
Actual calculations by KHAN = 9.33
28
Q

If 0.03 mol of HCl is added to a 0.50 L of the buffer solution (0.24 M NH3 and 0.20 M NH4Cl) what is the resulting pH?

A

The problem gives you a substance in mols. It asks for a pH which can only be found in molarity
Therefore 0.03mol/0.5L = 0.06M HCl

In adding more acid into the buffer system, the strong acid will dissociate into completion. This means HCl -> H+ and Cl-. H+ will producing H3O+ and of the weak acid and its ionization (therefore NH3 and NH4+) the H3O+ will interact with NH3. This push the reaction in favor of decreasing the ammonia into NH4

In doing the ICE table for any additions, make sure to add in the adjusted compound, in this case H3O+

NH3 + H2O ⇔ NH4+ + H2O-
I 0.24 0.06 0.2
C -0.06 -0.06 +0.06 0
E 0.18 0.26

pH = pKa + log[A-]/[HA]
= 9.5 + log[0.18]/[0.26]
= 9.5 - 0.16 = 9.34
Actual calculations from KHAN = 9.25 - 0.16 = 9.09

Buffer is amazing at resisting large changes in pH even though we added in quite a lot of acid into the solution! Buffers are amazing

29
Q

BF3 is added into an acid solution. What do you expect to happen in this reaction?

A

Ex: BF3 + H+
In reactions in which the electron is transferred can form bonds between the 2 molecules of focus. This bond formed through this electron transfer is a coritcoCOVALENT bond usually
Cortico because BOTH electrons come from a single atom! - The H. The H donates both electrons into bond making
This is a broad definition and can include both Arrhenius and Bronsted Lowry definitions

30
Q

Compare and contrast 3 definitions for acids and bases

A

Arrhenius: acids gives off H+ in solution; bases gives off OH- in solution
Bronsted-Lowry: Acids gives off H+ in solution; bases accepts H+ from solution
Lewis: Acids gives off e- pairs in solution; bases accepts e- pairs from solution

31
Q

Utilizing the Arrhenius Acid naming trends, predict the acid formula and name for the following anion: MnO4- and TiO32-

A

MnO4- => HMnO4- => permanganate

TiO32- => HTiO3- => Titanic Acid

32
Q

Utilizing the Arrhenius Acid naming trends, predict the acid formula and name for the following anion: I- and IO4-

A

I- => HI => Hydroiodic acid

IO4- => HIO4- => periodic acid

33
Q

[H+] = 0.001. What is the pH?

A

pH ~ 3

0.001 => 10-3 => log(10-3) = 3 - 0.1 = 2.9

34
Q

What is pKb when Kb=1.0x10-12?

A

~12

log(1.0x10-12) = 12 - 0.1 = 11.9

35
Q

If the Ka of an acid is 1.8 x 10-5, then what is the pKa?

A

Kaplan shortcut: 5 - 0.18 = 4.82

Actual value = 4.74

36
Q

Calculate the pH of a 1x10-8M solution of HCl.

A

pH = -log[H]
Remember that HCl is a strong acid and dissociated completely!
Remember that using the concentration in this case to calculate for the pH is not feasible because in this case, we will get that the pH = ~8, but that is not an acidic solution!
We have to take into account of the autoionization of water in this case because this value of 1x10-7M of H3O+ by autoionization is far larger than the concentration of the strong acid: 1x10-8M
Therefore summating this up and taking this into account in Kw = [x+10-8M][x] = 10-14
The mathematical equation of this formula would require the quadratic formula. But one thing to note is that when the strong bases and acids have less concentration compared to the concentration of H3O+ and OH-, then the pH or pOH will be very close to neutral

37
Q

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

A

Perchloric acid is a strong acid therefore complete dissociation of this creates 10 M of H3O+M. And because this value is far greater than the autoionization concentration of H3O+, this can be ignored
pH = -log[H3O+] = -log[10M] = -1
This is a good reminder that the pH scale does not end at 0! There are negatives as well! And this also means that the pH scale does not end at 14 either!

38
Q

Explain why the dissociation of bicarbonate acid into carbonate is used a buffer system in the body. Use the significance of conjugate acids and bases for your explanation.

A

HCO3- (aq) + H2O (l) ⇔ CO32-(aq) + H3O+(aq)
Conjugate acids and bases are of weak acids (like Bicarbonate HCO3-) are also weak as well! While this conjugate base is a weak, it has the reverse reaction is far more thermodynamically favored, which makes it a great buffer system in the body as it has the ability to decrease the H3O+ in the bloodstream

39
Q

What species are considered the equivalents for acids and bases, respectively?

A

Acids - H3O+

Bases - OH-

40
Q

Calculate the normality of 2M Al(OH)3

A

Al(OH)3 is a polyvalent base and potentially produces OH- with each dissociation. At full dissociation completion, 3 OH equivalents are produced
2M (3 equivalents) = 6N

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