Chem Lab C

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Chem Lab C

Post by N0sm4 » December 15th, 2020, 6:44 pm

Ok so let’s get this question marathon started :D

So I am going to start off with a relatively easy question

1. If you have 30 mL of 0.2 M NaOH, how many liters of 0.3 M HNO3 is required to turn the phenol red indicator in the solution red.
Last edited by N0sm4 on December 15th, 2020, 6:49 pm, edited 1 time in total.

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Re: Chem Lab C

Post by Mayur917 » December 15th, 2020, 7:55 pm

N0sm4 wrote:
December 15th, 2020, 6:44 pm
Ok so let’s get this question marathon started :D

So I am going to start off with a relatively easy question

1. If you have 30 mL of 0.2 M NaOH, how many liters of 0.3 M HNO3 is required to turn the phenol red indicator in the solution red.
Strong Acid + Strong Base » Neutral solution + H2O
HNO3 (aq) + NaOH (aq) » NaNO3 (aq) + H2O (l)

Molar ratios are all 1 so we can proceed with M1V1 = M2 V2

(0.2 M) (0.03 L) = (0.3) (V2)
0.02 L = V2
20 mL are required to turn the phenol red indicator into the solution red.
Is that the correct answer?
Last edited by Mayur917 on December 16th, 2020, 3:31 pm, edited 3 times in total.
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Re: Chem Lab C

Post by N0sm4 » December 16th, 2020, 12:46 pm

Mayur917 wrote:
December 15th, 2020, 7:55 pm
N0sm4 wrote:
December 15th, 2020, 6:44 pm
Ok so let’s get this question marathon started :D

So I am going to start off with a relatively easy question

1. If you have 30 mL of 0.2 M NaOH, how many liters of 0.3 M HNO3 is required to turn the phenol red indicator in the solution red.
Strong Acid + Strong Base » Neutral solution + H2O
HNO3 (aq) + NaOH (aq) » NaNO3 (aq) + H2O (l)

Molar ratios are all 1 so we can proceed with M1V1 = M2 V2

(0.2 M) (0.03 L) = (0.3) (V2)
0.02 L = V2
20 mL are required to turn the phenol red indicator into the solution red.

Is that the correct answer?
Yeah and also a note for next time remember to put your answer in between answer tags or in a spoiler.

Code: Select all

[answer]Your answer/work here[/answer]
Your turn now.
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Re: Chem Lab C

Post by RiverWalker88 » April 2nd, 2021, 9:30 am

Revive:

For some odd reason, a geology undergrad finds themselves in the storage closet of a chemistry lab at 1:30 AM. Not paying any attention, they knock over a 2.50 L bottle of HCl, that we just happen to somehow know had a pH of -1.204. Panicking, they grab the nearest basic solution, which just happened to be 3.90 M sodium hydroxide, and emptied the entire bottle onto the floor with the HCl.
  1. Assuming they managed to exactly neutralize the solution, how many liters of sodium hydroxide was contained in the bottle?
  2. How many grams of table salt is the student going to have to clean up off of the floor?
  3. List one laboratory safety procedure that this student did not follow.
Because this is chemistry, significant figures do matter.

Edit: Updated the last question to be more chemistry-like. It was originally asking for a moral of the story.
Last edited by RiverWalker88 on April 2nd, 2021, 9:37 am, edited 1 time in total.
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Re: Chem Lab C

Post by Jehosaphat » April 12th, 2021, 9:31 am

RiverWalker88 wrote:
April 2nd, 2021, 9:30 am
Revive:

For some odd reason, a geology undergrad finds themselves in the storage closet of a chemistry lab at 1:30 AM. Not paying any attention, they knock over a 2.50 L bottle of HCl, that we just happen to somehow know had a pH of -1.204. Panicking, they grab the nearest basic solution, which just happened to be 3.90 M sodium hydroxide, and emptied the entire bottle onto the floor with the HCl.
  1. Assuming they managed to exactly neutralize the solution, how many liters of sodium hydroxide was contained in the bottle?
  2. How many grams of table salt is the student going to have to clean up off of the floor?
  3. List one laboratory safety procedure that this student did not follow.
Because this is chemistry, significant figures do matter.

Edit: Updated the last question to be more chemistry-like. It was originally asking for a moral of the story.
I think this is as straight forward as I'm thinking, but here goes nothing.
 
a. 10.3 liters of NaOH - (2.50 L HCl x 10^1.204) / 3.90 M NaOH = 10.3 L NaOH
b. 2350 grams of NaCl - 3.90 M NaOH x 10.3 L NaOH = 40.17 mol NaOH x 1 mol NaCl = 40.17 mol NaCl x 58.44 g/mol = 2350 grams
c. They were most likely  not in full lab gear required for this circumstance (Long sleeves/pants, splash googles, closed toe shoes), that are most certainly needed in this situation.
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Re: Chem Lab C

Post by RiverWalker88 » April 14th, 2021, 6:29 am

Jehosaphat wrote:
April 12th, 2021, 9:31 am
RiverWalker88 wrote:
April 2nd, 2021, 9:30 am
Revive:

For some odd reason, a geology undergrad finds themselves in the storage closet of a chemistry lab at 1:30 AM. Not paying any attention, they knock over a 2.50 L bottle of HCl, that we just happen to somehow know had a pH of -1.204. Panicking, they grab the nearest basic solution, which just happened to be 3.90 M sodium hydroxide, and emptied the entire bottle onto the floor with the HCl.
  1. Assuming they managed to exactly neutralize the solution, how many liters of sodium hydroxide was contained in the bottle?
  2. How many grams of table salt is the student going to have to clean up off of the floor?
  3. List one laboratory safety procedure that this student did not follow.
Because this is chemistry, significant figures do matter.

Edit: Updated the last question to be more chemistry-like. It was originally asking for a moral of the story.
I think this is as straight forward as I'm thinking, but here goes nothing.
 
a. 10.3 liters of NaOH - (2.50 L HCl x 10^1.204) / 3.90 M NaOH = 10.3 L NaOH
b. 2350 grams of NaCl - 3.90 M NaOH x 10.3 L NaOH = 40.17 mol NaOH x 1 mol NaCl = 40.17 mol NaCl x 58.44 g/mol = 2350 grams
c. They were most likely  not in full lab gear required for this circumstance (Long sleeves/pants, splash googles, closed toe shoes), that are most certainly needed in this situation.
Yep, that's it. Your turn!
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Re: Chem Lab C

Post by Jehosaphat » April 14th, 2021, 7:20 pm

You are tritrating 40 mL of 0.20 M acetic acid with 0.25 M NaOH.

1. Describe the materials that are required to complete this experiment.
2. Calculate the initial pH of the solution.
3. Calculate the pH after 10 mL of NaOH is added.
4. Calculate the pH of the solution at the equivalence point.

Bonus: What about the structure of acetic acid causes it to behave as an acid?
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Re: Chem Lab C

Post by dxu46 » April 18th, 2021, 11:56 am

Jehosaphat wrote:
April 14th, 2021, 7:20 pm
You are tritrating 40 mL of 0.20 M acetic acid with 0.25 M NaOH.

1. Describe the materials that are required to complete this experiment.
2. Calculate the initial pH of the solution.
3. Calculate the pH after 10 mL of NaOH is added.
4. Calculate the pH of the solution at the equivalence point.

Bonus: What about the structure of acetic acid causes it to behave as an acid?
1. beaker, buret, volumetric pipet, stir plate/bar, indicator (usually phenolphthalein), pH probe or pH paper
2. pH = -log(sqrt(1.8*10^-5 * 0.25 M)) = 2.72
3. adding 0.0025 mol base to 0.008 mol acid, ICE yields 0.0055 mol acid and 0.0025 mol conjugate base, creating a buffer. Henderson-Hasselbalch yields: pH = pKa + log([A-]/[HA]), pH = 4.74 + log(0.0025/0.0055) = 4.40
4. mol acid = mol base --> = mol conjugate base, so 0.008 mol CH3COO-, total volume = 72 mL. ICE yields (x^2/0.111 = 1/Ka), x = 7.9*10^-6 M, which is [OH-], leading pH to be 14 - pOH = 8.90

Bonus: the O-H bond in the carboxyl group is a polar covalent bond that allows acetic acid to ionize and donate the proton, behaving as an acid. 

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Re: Chem Lab C

Post by Jehosaphat » April 19th, 2021, 6:35 am

dxu46 wrote:
April 18th, 2021, 11:56 am
Jehosaphat wrote:
April 14th, 2021, 7:20 pm
You are tritrating 40 mL of 0.20 M acetic acid with 0.25 M NaOH.

1. Describe the materials that are required to complete this experiment.
2. Calculate the initial pH of the solution.
3. Calculate the pH after 10 mL of NaOH is added.
4. Calculate the pH of the solution at the equivalence point.

Bonus: What about the structure of acetic acid causes it to behave as an acid?
1. beaker, buret, volumetric pipet, stir plate/bar, indicator (usually phenolphthalein), pH probe or pH paper
2. pH = -log(sqrt(1.8*10^-5 * 0.25 M)) = 2.72
3. adding 0.0025 mol base to 0.008 mol acid, ICE yields 0.0055 mol acid and 0.0025 mol conjugate base, creating a buffer. Henderson-Hasselbalch yields: pH = pKa + log([A-]/[HA]), pH = 4.74 + log(0.0025/0.0055) = 4.40
4. mol acid = mol base --> = mol conjugate base, so 0.008 mol CH3COO-, total volume = 72 mL. ICE yields (x^2/0.111 = 1/Ka), x = 7.9*10^-6 M, which is [OH-], leading pH to be 14 - pOH = 8.90

Bonus: the O-H bond in the carboxyl group is a polar covalent bond that allows acetic acid to ionize and donate the proton, behaving as an acid. 
Everything looks good to me! On number one you used the molarity of NaOH instead of the molarity of acetic acid, but you got the same answer anyways. Your turn!
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Re: Chem Lab C

Post by dxu46 » April 20th, 2021, 9:00 pm

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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