Calculate Acetic Acid Adsorbed by Activated Charcoal

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of acetic acid adsorbed by activated charcoal. Participants explore the relationship between normality, molarity, and the calculation of moles before and after adsorption, as well as the implications of volume changes in the solution.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant initially attempts to find the volume of the acetic acid solution but later realizes that it may not be necessary for the calculation.
  • Another participant suggests calculating the number of moles of acetic acid before and after adsorption to determine how much was adsorbed by the charcoal.
  • There is a discussion about the n factor for acetic acid being 1, which implies that its molarity is the same as its normality.
  • Participants calculate the number of moles in the solution before and after adsorption using the provided concentrations and volumes.
  • One participant expresses confusion about the assumption that the volume does not change with the addition of solutes, while another clarifies that in many cases, the change can be neglected.
  • Steps for calculating the moles of acetic acid adsorbed are outlined, leading to a discrepancy between the participant's calculation and the textbook answer.
  • Participants discuss the negligible volume change when small amounts of solute are removed from a solution, emphasizing the importance of understanding when to keep volume constant in calculations.

Areas of Agreement / Disagreement

Participants generally agree on the method of calculating moles before and after adsorption, but there is some disagreement regarding the treatment of volume changes and the resulting calculations. The discussion remains unresolved regarding the correct interpretation of the textbook answer.

Contextual Notes

Participants note that the assumption of constant volume may not hold in cases of significant concentration changes, but in this scenario, the volume change is considered negligible. There are also unresolved aspects regarding the correct application of the calculations leading to differing answers.

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Homework Statement



(Q.) 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :

Homework Equations

The Attempt at a Solution


I thought of finding the volume first by using the formulae :N1V1=N2V2 and then get the volume but I realized that finding volume won't help so I am out of ideas
 
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The volume doesn't change.

How many moles of the acetic acid before adding the charcoal, how many moles after adsorption?
 
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the n factor for acetic acid is 1 so it's molarity will be same (o.o6M). and from the formulae ,we can calulate the moles but I don't know how to calculate moles after adsorption
 
You have 50 mL of a 0.042 M solution, how do you calculate number of moles in such case?

What is the definition of the molar concentration?
 
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Borek said:
You have 50 mL of a 0.042 M solution, how do you calculate number of moles in such case?
Molarity X Volume=0.042 X 0.05
 
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Borek said:
The volume doesn't change.

Doesn't volume increase with addition of solutes
 
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Hydrous Caperilla said:
Doesn't volume increase with addition of solutes

In general you are right, it does. But in most cases change can be neglected.

Please pay attention to what you post, each your post contains unnecessary multiple quotes (I will remove them now). Use "preview" button to see what it will look like. You can also edit your posts for a short period of time.
 
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Borek said:
In general you are right, it does. But in most cases change can be neglected.

Please pay attention to what you post, each your post contains unnecessary multiple quotes (I will remove them now). Use "preview" button to see what it will look like. You can also edit your posts for a short period of time.

So I want to post the steps I have understood

Step 1 Before Adsorption

No of moles of acetic acid in the solution=50X0.6=3milli moles(n factor of acetic acid is 1 so Normality=Molarity)

Step 2
No of mole of acetic acid after adsorption=50(volume is not changing which I don't understand)X0.042=2.1 millimoles

Step 3
Moles adsorbed= O.9millimoles

Step 4

Weight of acetic acid=0.9X60/1000=54mg
My textbook gives the answer as 18mg...what did I do wrong now and can you tell me more about when to keep volume constant
 
Hydrous Caperilla said:
Weight of acetic acid=0.9X60/1000=54mg

That's correct.

Hydrous Caperilla said:
My textbook gives the answer as 18mg

Looks OK.

Hydrous Caperilla said:
what did I do wrong

Reread the question, you have missed something.

Hydrous Caperilla said:
when to keep volume constant

Almost always, only when the concentration changes are huge it really matters.

0.06 M acetic acid contains 3.6 g of acetic acid per liter.
0.042 M acetic acid contains 2.5 g of acetic acid per liter.
The difference is 1.1 g. Imagine removing the acid from the solution - you are removing 1 g from 1000 g, so you can expect volume change in the 0.1% range at best. Negligible.
 
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  • #10
I get it now.
I forgot to divide it by the mass of charcoal so it makes sense now.
Thanks
 

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