Check Concentration Calculation (Very Wordy As For Coursework)

In summary, My friend's solution has a concentration of 0.83moldm-3 while my solution has a concentration of 0.005moldm-3.
  • #1
Kevlarji
6
0

Homework Statement



You are provided with a solution of hydrogen peroxide which is known to be approximately ‘100-volume’. Plan two experiments that would enable you to determine the exact concentration of H2O2 in mol dm-3, in this solution

Homework Equations



2H2O2 (aq) ----------------> O2 (g) + 2H2O (l)

The Attempt at a Solution



My friends solution

The term ‘100-volume’ means that for every 1cm3 of H2O2, 100cm3 of oxygen gas will be liberated. As this volume of gas is not ideal for a chemical experiment, the H2O2 will have to be diluted by a factor of 10 to give is 10-volume H2O2 instead. This will be done by the following steps:

1) Accurately measure 25cm3 of ‘100-volume’ H2O2 using a graduated pipette and empty into a 250cm3 conical flask.
2) Fill the conical flask up to 250cm using distilled water

The oxygen gas will be collected and measured by using gas syringe. As a typical gas syringe holds about 100cm3, a desirable amount of oxygen to collect would be about 60cm3. 10-volume H2O2 liberates 10cm3 of oxygen for every 1cm3 of H2O2 and so therefore 6cm3 of H2O2 will be required. From this the concentration of H2O2 will be calculated as follows:

-1mole of gas = 24000cm3.
-As 60cm3 of gas is collected, moles = 60/24000 = 0.0025moles
-The stoichiometry of the equation states that for every mole of oxygen, there are twice as many of H2O2. Therefore moles of H2O2 = 2 x 0.0025 = 0.005moles
-concentration = moles/volume (in dm3); 6cm3 = 0.006dm3.
Concentration = 0.005/0.006 = 0.83moldm-3
-As the original concentration of H2O2 was diluted by a factor of 10, the original concentration is 10times as much as 0.83moldm-3 and so therefore is 0.83 x 10 = 8.3moldm-3.

My Solution

‘100-Volume’ means that 1cm3 of H2O2 will release 100cm3 of oxygen gas. Starting with a 100-Volume H2O2 would be unreasonable as a gas syringe can only hold up to 100cm3 of gas, therefore I am going to dilute the H2O2 by a factor of 10 making a ’10-Volume’ solution.
This will be done by taking 25cm3 of 100-Volume H2O2 into a bulb pipette then putting it into a 250cm3 conical flask and then adding 225cm3 of distilled water thus creating a ’10-Volume’ solution. This now means 1cm3 of H2O2 will release 10cm3 of oxygen gas.

Knowing a gas syringe can only hold up to 100cm3 of gas a good sample of gas to collect would be 70cm3 of oxygen. To get 70cm3 I would require 7cm3 of 10-Volume H2O2

With this information I can calculate the concentration of 100-Volume H2O2

Firstly I must convert 70cm3 to dm3 dividing through by 1000
70/1000=0.07 dm3

Secondly I must use the rule that one mole of any gas occupies 24dm3 at room temperature, this is to calculate the moles of O2.
0.07/24=0.00292 moles

I can now use this information to determine the moles of H2O2. Using the equation the moles of H2O2 to moles of O2 is 2:1 so
Moles of H2O2 = 0.00292 x 2 = 0.00583 moles

I can now calculate the concentration of 10-Volume H2O2.
Concentration = (Moles/Volume) mol dm-3
(0.00583)/0.07=0.083 mol dm-3

This is the concentration of the 10-Volume solution to get the solution of the 100-Volume solution I must times 0.083 by 10 which gives a concentration 0.83 dm-3 of 100-Volume H2O2.

So whos right me or my friend?
 
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  • #2
Kevlarji said:
(0.00583)/0.07=0.083 mol dm-3

What was the volume of the solution used?
 
  • #3
Ahh Thanks I was dividing by 0.07dm3 O2 not the 0.007dm3 of h2o2 T_T!
 

1. What is meant by "Check Concentration Calculation" in scientific research?

"Check Concentration Calculation" refers to the process of double-checking and verifying the accuracy of the concentration values calculated for a solution or sample in a scientific experiment. This is an important step in ensuring the reliability and validity of the results.

2. Why is it necessary to perform a check concentration calculation?

Performing a check concentration calculation is necessary to ensure that any errors in the original concentration calculations are identified and corrected. This helps to eliminate potential sources of error and increases the confidence in the experimental results.

3. How is a check concentration calculation typically performed?

A check concentration calculation is typically performed by using a different method or technique to determine the concentration of the same solution or sample. The results from the two methods are then compared, and any discrepancies are investigated and resolved.

4. What are the potential sources of error in concentration calculations?

Potential sources of error in concentration calculations include human error, equipment malfunctions, incorrect measurements, and contamination of the sample. These errors can lead to inaccurate results and can be identified and addressed through a check concentration calculation.

5. How can I ensure the accuracy of my check concentration calculation?

To ensure the accuracy of a check concentration calculation, it is important to follow proper experimental procedures and carefully record all measurements and calculations. It is also helpful to have a colleague or lab partner double-check the calculations for verification. Additionally, using multiple methods or techniques to determine the concentration can help to increase the accuracy of the results.

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