Help: Analytical Chem Tritration Problem

In summary, the conversation is about a back tritration problem involving a sample of spring water that was treated to convert iron present to Fe2+. The addition of K2Cr2O7 resulted in a reaction that was later back titrated with Fe2+ solution. The conversation also includes calculations and questions about the concentration of iron in the sample and the amount of Fe2+ required and added during the titration.
  • #1
bdanish
1
0
Can someone please help me with this back tritration problem!

a 100.0 ml sample of spring water was treated to convert any iron present to Fe2+ (AW 55.847) Addition of 25.00 ml of .002107 M K2Cr2O7 (FW294.185) resulted in the reaction:

6(Fe2+) + (Cr2O7)(2-) + 14H+ _____6(Fe3+) + 2(Cr3+) = 7H2O

the excess K2Cr2O7 was back titrated with 7.47 ml of .00979 M Fe2+ solution. Calculate the concentration (in ppm) of the iron in the sample (Recall: ppm Fe = #g Fe/10^6g sample).
 
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  • #2
How many moles of Fe2+ are required to completely react given amount of dichromate?

How many were added during titration?

How many reacted in the first stage?
 
  • #3



First, we need to determine the moles of K2Cr2O7 used in the initial reaction. This can be calculated by multiplying the volume (25.00 ml) by the molarity (.002107 M) and dividing by the FW (294.185 g/mol):

moles K2Cr2O7 = (25.00 ml)(.002107 M) / 294.185 g/mol = 0.0001788 moles

Next, we need to determine the moles of Fe2+ present in the sample. Since the reaction is 6:1, the moles of Fe2+ will be 6 times the moles of K2Cr2O7:

moles Fe2+ = 0.0001788 moles K2Cr2O7 x 6 = 0.0010728 moles

Now, we need to determine the moles of Fe2+ used in the back titration. This can be calculated by multiplying the volume (7.47 ml) by the molarity (.00979 M):

moles Fe2+ = (7.47 ml)(.00979 M) = 0.0731433 moles

Since the reaction is 1:1, the moles of Fe2+ used in the back titration is the same as the moles of Fe2+ present in the sample. Therefore, the concentration of Fe2+ in the sample is:

0.0731433 moles / 0.1000 L = 0.731433 M

To convert this to ppm, we need to multiply by the FW (55.847 g/mol) and divide by the sample size (100.0 ml) and multiply by 10^6:

ppm Fe = (0.731433 M)(55.847 g/mol) / (100.0 ml) x 10^6 = 4092.75 ppm

Therefore, the concentration of iron in the sample is 4092.75 ppm.
 

1. What is analytical chemistry?

Analytical chemistry is a branch of chemistry that focuses on the identification and quantification of chemical substances in various samples. It involves techniques such as titration, chromatography, and spectroscopy to analyze the composition and properties of substances.

2. What is a titration problem in analytical chemistry?

A titration problem in analytical chemistry involves the use of a known concentration of a substance (titrant) to determine the concentration of an unknown substance (analyte) in a sample. It is a common method used to determine the amount of a substance present in a solution.

3. How do you solve a titration problem?

To solve a titration problem, you need to first determine the volume and concentration of the titrant used. Then, you can use the known stoichiometry of the reaction between the titrant and analyte to calculate the concentration of the analyte. This can be done using the formula: concentration of analyte = (volume of titrant * concentration of titrant) / volume of analyte.

4. What is the purpose of using an indicator in titration?

An indicator is a substance that changes color when the endpoint of a titration is reached. It is used to visually determine when the reaction between the titrant and analyte is complete. This helps to ensure accurate and precise results in the titration process.

5. What are common sources of error in titration experiments?

Some common sources of error in titration experiments include human error in measuring volumes or concentrations, improper calibration of equipment, and impurities in the titrant or sample. It is important to carefully follow proper techniques and protocols to minimize these errors and obtain accurate results.

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