Chem Lab C

Test your knowledge of various Science Olympiad events.
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Jehosaphat
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Re: Chem Lab C

Post by Jehosaphat » April 21st, 2021, 11:42 am

1. Rank the following acids or bases based on increasing strength, and provide a justification based on molecular structure/bonding/etc. (i.e. no searching up Ka/Kb values):
a. HClO, HIO, HBrO, HFO
b. HClO, HClO2, HClO3, HClO4
c. NH3, NF3, NCl3, NBr3
d. NH3, PH3, AsH3, SbH3

2. Below is the fictional titration curve for the fictional acid, H2Dx:
Image
a. Why does the titration curve have two equivalence points?
b. Classify H2Dx as strong or weak, and explain why.
c. Estimate the pKa of HDx-. Why can't we estimate the pKa of H2Dx?
d. Can HDx- act as a base? Explain why using two equilibrium equations. What is this called when a substance can act as both an acid and a base?
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1a. HFO, HClO, HBrO, HIO. The strength of the acid is affected by the electronegativity of the attached halogen. Flourine has the strongest attraction to the H+ ion, so it is the weakest, and Iodine has the weakest attraction, so it is the strongest acid.
1b. HClO, HClO2, HClO3, HClO4. The more oxygens there are pulling the electrons away from the central chlorine atom, the weaker the attraction between Cl and H will be. 
1c. NF3, NCL3, NBr3, NH3. Flourine's electronegativity will pull the electrons away from nitrogen, giving the nitrogen a partial negative charge, making it much less likely to attract the hydrogen ion. In NH3, where the hydrogen have a much lower electronegativity, the nitrogen will have a partial negative charge, which attracts the hydrogen ion, making it a stronger base.
1d. SbH3, AsH3, PH3, NH3. The same rationale except it is for the central atom. Nitrogen has a stronger electronegativity, making it more partially negative. 

2. Your image didn't upload right, so I can't see it. 
Last edited by Jehosaphat on April 22nd, 2021, 4:50 am, edited 2 times in total.
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Re: Chem Lab C

Post by dxu46 » April 22nd, 2021, 10:31 am

Jehosaphat wrote:
April 21st, 2021, 11:42 am
1. Rank the following acids or bases based on increasing strength, and provide a justification based on molecular structure/bonding/etc. (i.e. no searching up Ka/Kb values):
a. HClO, HIO, HBrO, HFO
b. HClO, HClO2, HClO3, HClO4
c. NH3, NF3, NCl3, NBr3
d. NH3, PH3, AsH3, SbH3

2. Below is the fictional titration curve for the fictional acid, H2Dx:
Image
a. Why does the titration curve have two equivalence points?
b. Classify H2Dx as strong or weak, and explain why.
c. Estimate the pKa of HDx-. Why can't we estimate the pKa of H2Dx?
d. Can HDx- act as a base? Explain why using two equilibrium equations. What is this called when a substance can act as both an acid and a base?
Top
1a. HFO, HClO, HBrO, HIO. The strength of the acid is affected by the electronegativity of the attached halogen. Flourine has the strongest attraction to the H+ ion, so it is the weakest, and Iodine has the weakest attraction, so it is the strongest acid.
1b. HClO, HClO2, HClO3, HClO4. The more oxygens there are pulling the electrons away from the central chlorine atom, the weaker the attraction between Cl and H will be. 
1c. NF3, NCL3, NBr3, NH3. Flourine's electronegativity will pull the electrons away from nitrogen, giving the nitrogen a partial negative charge, making it much less likely to attract the hydrogen ion. In NH3, where the hydrogen have a much lower electronegativity, the nitrogen will have a partial negative charge, which attracts the hydrogen ion, making it a stronger base.
1d. SbH3, AsH3, PH3, NH3. The same rationale except it is for the central atom. Nitrogen has a stronger electronegativity, making it more partially negative. 

2. Your image didn't upload right, so I can't see it. 
Your answers are correct for question 1. I would've talked more about the definition of an acid for parts a and b and why a weaker O-H bond means stronger acid, but you were correct. (also F is spelled Fluorine not Flourine :D )

EDIT: You probably didn't mean this, but you said "central chlorine atom" and "attraction between Cl and H", when chlorine should be replaced with oxygen (the hydrogen in an oxoacid is always attached to an oxygen).

Sorry the image didn't upload, it appears fine on my end (and I think you can always right click on the image to open it in a new tab), but anyways I'll attach my solution:
a. H2Dx is a diprotic acid, which means that each equivalence point represents the loss of one proton.
b. Strong, because the titration curve for the first hydrogen doesn't have the characteristic curve of a weak acid-strong base titration (i.e., no half equivalence point)
c. The pKa is estimated to be ~6.1, because it is where the concentrations of Dx2- and HDx- are equal, and thus is the value of the pKa, given by the Henderson-Hasselbalch equation. Because H2Dx is a strong acid, the Ka is much greater than 1, it has a low buffering capacity, and cannot be estimated based on this graph.
d. Yes; amphoteric. I'm too lazy to write the equilibrium equations out.
Your turn!
Last edited by dxu46 on April 22nd, 2021, 10:49 am, edited 1 time in total.
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Re: Chem Lab C

Post by Jehosaphat » April 22nd, 2021, 12:50 pm

Alright, here's your situation.

You have a liter of 1 M NaOH at your disposal. There is a beaker filled with 55 mL of 0.3 M ascorbic acid (Ka1=8.0x10-5, Ka2=1.6x10-12).
1. What is the percent ionization of the ascorbic acid?
2. What is the pH of the ascorbic acid solution?
3. How many mL of NaOH must you add to neutralize the acid?
5. Describe how you would prepare a buffer using the materials given.
6. What is the pH of the buffer solution?
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Re: Chem Lab C

Post by dxu46 » April 23rd, 2021, 11:10 am

Jehosaphat wrote:
April 22nd, 2021, 12:50 pm
Alright, here's your situation.

You have a liter of 1 M NaOH at your disposal. There is a beaker filled with 55 mL of 0.3 M ascorbic acid (Ka1=8.0x10-5, Ka2=1.6x10-12).
1. What is the percent ionization of the ascorbic acid?
2. What is the pH of the ascorbic acid solution?
3. How many mL of NaOH must you add to neutralize the acid?
5. Describe how you would prepare a buffer using the materials given.
6. What is the pH of the buffer solution?
1. %I = [H+]/[HA]. [H+] = sqrt(Ka1 * 0.3 M) = 0.0049 M, %I = 0.016 = 1.6% ionized
2. Using ICE --> (x^2/0.3-x = Ka1, x = 0.0049 M). Now we have an ascorbic acid/biascorbate buffer, so Henderson-Hasselbalch applies, pH = -log(Ka1) + log(0.0049/(0.3-0.0049)) = 2.31 (also, -log(0.0049) = 2.31)
3. [H+] in ascorbic acid = 0.0049 M = 2.7*10^-4 mol H+, divide by 1 M OH- = 2.7*10^-4 L = 0.27 mL NaOH
5. To maximize the buffering capacity, the concentration of biascorbate must be the same as ascorbic acid, so when [A-] = [HA] = 0.15 M. Enough base must be added to increase the concentration of A- to 0.15 M. Assuming that standard lab materials are readily available, use a volumetric pipet to measure out exactly that amount of base, then dispense it into the beaker, and mix.
6. At maximum buffering capacity (Henderson-Hasselbalch, when [A-] = [HA]), pH = pKa, so pH = -log(Ka1) = 4.10

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Re: Chem Lab C

Post by Jehosaphat » April 23rd, 2021, 3:32 pm

dxu46 wrote:
April 23rd, 2021, 11:10 am
Jehosaphat wrote:
April 22nd, 2021, 12:50 pm
Alright, here's your situation.

You have a liter of 1 M NaOH at your disposal. There is a beaker filled with 55 mL of 0.3 M ascorbic acid (Ka1=8.0x10-5, Ka2=1.6x10-12).
1. What is the percent ionization of the ascorbic acid?
2. What is the pH of the ascorbic acid solution?
3. How many mL of NaOH must you add to neutralize the acid?
5. Describe how you would prepare a buffer using the materials given.
6. What is the pH of the buffer solution?
1. %I = [H+]/[HA]. [H+] = sqrt(Ka1 * 0.3 M) = 0.0049 M, %I = 0.016 = 1.6% ionized
2. Using ICE --> (x^2/0.3-x = Ka1, x = 0.0049 M). Now we have an ascorbic acid/biascorbate buffer, so Henderson-Hasselbalch applies, pH = -log(Ka1) + log(0.0049/(0.3-0.0049)) = 2.31 (also, -log(0.0049) = 2.31)
3. [H+] in ascorbic acid = 0.0049 M = 2.7*10^-4 mol H+, divide by 1 M OH- = 2.7*10^-4 L = 0.27 mL NaOH
5. To maximize the buffering capacity, the concentration of biascorbate must be the same as ascorbic acid, so when [A-] = [HA] = 0.15 M. Enough base must be added to increase the concentration of A- to 0.15 M. Assuming that standard lab materials are readily available, use a volumetric pipet to measure out exactly that amount of base, then dispense it into the beaker, and mix.
6. At maximum buffering capacity (Henderson-Hasselbalch, when [A-] = [HA]), pH = pKa, so pH = -log(Ka1) = 4.10
Everything looks good to me! For number 4 you could've also just said add half of the volume of NaOH needed to neutralize the solution , but your answer is better. I really meant to change the molarity of the NaOH to make it a bit more difficult, but I forgot.
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Re: Chem Lab C

Post by dxu46 » April 28th, 2021, 6:40 am

Let's shift over to some aqueous solutions questions:
1. What is a colligative property? List three examples.
2. 63.0g of NaCl is added to 0.300L of water at 23.0°C. What is the freezing point of the solution? The Kf for water is 1.86°C/m. Assume the density of water is 1.00g/mL at this temperature.

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