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Chemistry 101 COPY > Practice Questions > Flashcards

Practice Questions Flashcards

1
Q

What is the pH of the solution when 0.1 M of acetic acid and 0.05 M of sodium acetate are mixed, pKa=4.76?

A

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

∴pH=4.76+log[0.05]/[0.1]

∴pH=4.76+log0.5

∴pH=4.76|(-0.3)|

∴pH = 4.46

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2
Q

Calculate the pH of a buffer containing 0.05 M pyruvic acid and 0.06M sodium pyruvate, Ka of pyruvic acid = 3.2x10-3

A

The first step is to covert Ka to pKa:

pKa = -logKa

∴​pKa = -log(3.2x10-3)

∴​pKa = 2.49

→pH = pKa + log[A-]/[HA]

∴​pH = 2.49 + log[0.06]/[0.05]

∴​pH = 2.49 + log[1.2]

∴​pH = 2.49 + 0.079

∴​pH = 2.57

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3
Q

Prepare a buffer solution with a pH of 4.0

A

First step: Choose the buffer:

e.g. Lactic acid (pKa = 3.85)

Second step: Find ratio of buffer components:

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

[A-]/[HA] = 10(pH - pKa)

[A-]/[HA] = 10(4 - 3.85)

[A-]/[HA] = 1.41

→1.41 moles of sodium lactate: 1 mole of lactic acid

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4
Q

What is the pH of a solution of 0.04 M HNO3?

A

HNO3 is a strong acid

∴pH = -log[H+]

∴pH = -log(0.04)

∴pH = 1.40

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5
Q

What is the pH of a solution of 0.1 M HClO4?

A

HClO4 (perchloric acid) is a strong acid

∴pH = -log[H+]

∴pH = -log(0.1)

∴pH = 1

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6
Q

What is the pKa of C6H5CO2H if its Ka is 6.45x10-5?

A

pKa = -log(Ka)

∴pKa= -log(6.45x10-5)

∴pKa = 4.19

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7
Q

What is the Ka of triethylamine if its pKa is 11.01?

A

pKa = -log(Ka)

∴​Ka = 10-pKa

∴​Ka = 10-11.01

∴​Ka = 9.77x10-12

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8
Q

What is the pH of a solution of 0.01 M C6H5CO2H, pKa(C6H5CO2H) = 4.05?

A

Ka=10-pKa

∴​Ka=10-4.05

∴Ka=8.91x10-5

→[H+]=√(Ka x [HA])

∴[H+]=√(8.91x10-5x0.01)

∴[H+]=9.44x10-4

pH = -log[H+]

∴pH = -log(9.44x10-4)

∴pH = 3.03

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9
Q

What is the pH of a solution of 0.5 M trimethylamine, pKa(NMe3) = 9.81?

A

Ka = 10-pKa

∴Ka = 10-9.81

∴Ka = 1.55 x 10-10

→[H+] = √(Ka x [HA])

∴[H+] = √(1.55 x 10-10 x 0.5)

∴[H+] = 8.80 x 10-6​

→pH = -log[H+]

∴pH = -log(8.80 x 10-6)

∴pH = 5.06

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10
Q

Prepare a 2 L buffer solution with a pH of 9.50

A

Buffer to be used: Phenol: pKa = 9. 89

Using the Henderson-Hasselbalch equation:

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

∴9.5 = 9.89 + log([A-]/[HA])

∴-0.39 = log([A-]/[HA])

∴0.41 = ([A-]/[HA])

∴dissolve 0.41 moles of sodium phenoxide and 1 mole of phenol in water​

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11
Q

Prepare a 100 ml buffer solution with a pH of 7.5.

A

Buffer to be used: HEPES pKa = 7.66​

Using the Henderson-Hasselbalch equation:

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

∴7.5 = 7.66 + log([A-]/[HA])

∴​-0.16 = log([A-]/[HA])​

∴​0.69 = ([A-]/[HA])

∴​Dissolve 0.69 moles of sodium salt of HEPES and 1 mole of HEPES in water

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12
Q
A
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13
Q

Does this structure exist? Why/Why Not?

A

This structure does not exist. As an amino acid consists of an acid and a base, it will always be protonated in one area, and thus cannot be uncharged.

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14
Q
A
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15
Q

The pKa of the amino acid carboxylic acid is very low compared with acetic acid (pKa = 4.75); why?

A

The amino group is electron-withdrawing due to its positive charge, enhancing the acidity of the carboxylic acid and hence reducing pKa

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16
Q
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17
Q
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18
Q
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19
Q

Explain why the amine pKa of proline differs significantly from glycine?

A
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20
Q

Locate the amide bond(s) in this molecule

A
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21
Q

Build the Ser-Ala di-peptide from:

Serine: NH2CH(CH2OH)COOH and Alanine: NH2CH(CH3)COOH

A
22
Q
A
23
Q

Show an anti-parallel Beta Sheet

A
24
Q

Show a parallel Beta Sheet

A
25
Q

Which is preferred: parallel sheets or anti-parallel sheets?

A

Anti-parallel sheets due to minimised steric hindrance

26
Q
A

The trans isomer is the most stable: it has no (or less) steric hindrance

27
Q

Label the interactions shown in this diagram:

A
28
Q
A
29
Q
A
30
Q

What is the degree of ionisation if [A-]=1M and [HA]=0.1M?

A

%A-=[A-]/([HA]+[A-])x100

∴%A-=1/(0.1+1)x100

∴%A-=90.9%

31
Q

If [A-]=1M, [HA]=0.1M and %A-=90.9%, What is the pH of this solution If pKa (HA)=5.6?

A

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

∴pH=pKa+log[1]/[0.1]

∴pH=pKa+log10

∴pH=pKa+1

∴pH=5.6+1

∴pH=6.6

32
Q

If [A-]=1M, [HA]=0.1M, %A-=90.9%, pKa (HA)=5.6, and the pH of solution is 6.6, what can we conclude?

A

At pH 6.6, 90.9% of the compound is as A-

33
Q

What is the degree of ionisation of HA at pH 4.1 if the pKa of HA is 5.2?

A

pKa-pH=5.2-4.1=1.1

∴%A-= 1/(101.1+1)x100

∴%A-= 1/(12.58+1)x100

∴%A-= 1/13.58x100

∴%A-=7.36%

∴HA is present in 92.64%

34
Q

What is the degree of ionisation of benzoic acid at pH 3.1 if its pKa is 4.19?

A

%A- =1/(10(pka-pH)+1)x100

∴%A-=1/(101.09+1)x100

∴%A-=1/(12.30+1)x100

∴%A-=7.5%

35
Q

What is the percentage of benzoic acid present at pH 6.1 if its pKa is 4.19?

A

%HA=1/(10(pH-pka)+1)x100

∴%HA=1/(10(6.1-4.19)+1)x100

∴%HA=1/(101.91)+1)x100

∴%HA=1/(81.28+1)x100

∴%HA=1/82.28+1x100

∴%HA=1.21%

36
Q
A
37
Q

If the amine in amphetamine has a pKa = 9.8 is the drug active in the blood (pH 7.4) in its anionic or cationic form? Explain your answer.

A

Cationic, as pH < pKa

38
Q
A
39
Q

Determine the degree of ionisation if [A-]=2M and [HA]=0.02M

A

%A-=[A-]/([HA]+[A-])x100

%A-=[2]/([0.02]+[2])x100

%A-=99%

40
Q

Determine the degree of ionisation of HA at pH 3 if the pKa of HA is 4.7

A

%A-=1/(10(pka-pH)+1)x100

%A-=1/(10(4.7-3)+1)x100

%A-=1.96%

41
Q

Determine the degree of ionisation of HA at pH 8 if the pKa is 7.5.

A

%A-=1/(10(pka-pH)+1)x100

%A-=1/(10(7.5-8)+1)x100

%A-=75.97%

42
Q

Warfarin has a pKa=5, Ka=10-5M.

(a) What is the extent of its ionisation at pH 2?
(b) What is the extent of its ionisation at pH 7.4?

A

(a): %A- =1/(10(pka-pH)+1)x100

∴%A-=1/(10(5-2)+1)x100=0.10%

(a): %A- =1/(10(pka-pH)+1)x100

∴%A-=1/(10(5-7.4)+1)x100=99.60%

43
Q

Phenytoin has a pKa 8.3, Ka = 5.01x10-9M

(a) What is the extent of its ionisation at pH 2?
(b) What is the extent of its ionisation at pH 7.4?

A

(a) %A-=1/(10(8.3-2)+1)x100=0.00005%
(b) %A- =1/(10(7.4-2)+1)x100=11.2%

44
Q
A
45
Q
A
46
Q

Using straight chain nomenclature draw all the possible stereoisomers for erythose and describe the relationship

A
47
Q

What can be said about linear formations of sugars?

A

All sugars spontaneously cyclise to the more stable pyranose form, we can never isolate the linear structure in solution

48
Q

What do sugars form when they cyclise?

A

A new functional group: a hemiacetal

49
Q
A
50
Q

Show this sugar in its cyclic arrangement

A

Chemistry 101 COPY (32 decks)

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