Practice Questions Flashcards
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?
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
Calculate the pH of a buffer containing 0.05 M pyruvic acid and 0.06M sodium pyruvate, Ka of pyruvic acid = 3.2x10-3
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
Prepare a buffer solution with a pH of 4.0
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
What is the pH of a solution of 0.04 M HNO3?
HNO3 is a strong acid
∴pH = -log[H+]
∴pH = -log(0.04)
∴pH = 1.40
What is the pH of a solution of 0.1 M HClO4?
HClO4 (perchloric acid) is a strong acid
∴pH = -log[H+]
∴pH = -log(0.1)
∴pH = 1
What is the pKa of C6H5CO2H if its Ka is 6.45x10-5?
pKa = -log(Ka)
∴pKa= -log(6.45x10-5)
∴pKa = 4.19
What is the Ka of triethylamine if its pKa is 11.01?
pKa = -log(Ka)
∴Ka = 10-pKa
∴Ka = 10-11.01
∴Ka = 9.77x10-12
What is the pH of a solution of 0.01 M C6H5CO2H, pKa(C6H5CO2H) = 4.05?
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
What is the pH of a solution of 0.5 M trimethylamine, pKa(NMe3) = 9.81?
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
Prepare a 2 L buffer solution with a pH of 9.50
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
Prepare a 100 ml buffer solution with a pH of 7.5.
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
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/936/q_image_thumb.png?1509600366)
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/936/a_image_thumb.png?1509600607)
Does this structure exist? Why/Why Not?
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/938/q_image_thumb.png?1509601056)
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.
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/939/q_image_thumb.png?1509601788)
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/939/a_image_thumb.png?1509601708)
The pKa of the amino acid carboxylic acid is very low compared with acetic acid (pKa = 4.75); why?
The amino group is electron-withdrawing due to its positive charge, enhancing the acidity of the carboxylic acid and hence reducing pKa
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/941/q_image_thumb.png?1509604803)
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/941/a_image_thumb.png?1509604820)
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/943/q_image_thumb.png?1509605397)
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/943/a_image_thumb.png?1509605405)
Explain why the amine pKa of proline differs significantly from glycine?
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/946/a_image_thumb.png?1509606050)
Locate the amide bond(s) in this molecule
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/948/q_image_thumb.png?1509611814)
![](https://s3.amazonaws.com/brainscape-prod/system/cm/257/734/948/a_image_thumb.png?1509611889)