Determining the molality of a solution

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Discussion Overview

The discussion revolves around calculating the final molarity of a permanganate solution prepared by mixing manganese ions (Mn2+), phosphoric acid (H3PO4), and potassium periodate in a laboratory setting. Participants explore the implications of the initial concentrations and volumes of the reactants, as well as the stoichiometry of the resulting chemical reaction.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the molarity of the permanganate solution based on the initial concentration of Mn2+ and the final volume of the solution.
  • Another participant questions the initial conversion of concentration from ppm to molarity, suggesting it may have been done incorrectly.
  • There is a request for clarification on the number of moles of manganese and the stoichiometry of the reaction, specifically how many moles of permanganate are produced per mole of manganese.
  • A later reply indicates a correction in the conversion of concentration from micrograms per milliliter to molarity, providing a new value and reiterating the importance of considering the limiting reagent in the reaction.
  • Participants discuss the role of H3PO4 in the reaction, with one questioning whether it is involved in the stoichiometry of the final product.

Areas of Agreement / Disagreement

Participants express differing views on the accuracy of the concentration conversions and the stoichiometric relationships in the reaction. There is no consensus on the correct final molarity or the role of H3PO4 in the reaction.

Contextual Notes

Participants note potential issues with rounding errors and the use of different molar masses in their calculations. The discussion highlights the complexity of determining the final concentration due to the involvement of multiple reactants and the need for clarity on definitions and conversions.

georgiabrown
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Homework Statement



In a lab, 40.3 mL Mn2+ is mixed with 5 mL H3PO4. 0.4g of potassium periodate is added. The flask is heated and then diluted to 100mL. The final solution prepared is a permanganate solution. Calculate the final molarity of the solution.

C Mn2+ = 0.06061 M (60.61 micrograms/ mL)
C H3PO4 = 15 M
V Mn2+ = 40.3 mL
V H3PO4 = 5 mL
M (Mn2+) = 54.94 g/mol
V final = 100mL

Homework Equations



C1V1=C2V2

2Mn2+ + 5IO4- + 3H2O ---> 2MnO4- + 5IO3- + 6H+

The Attempt at a Solution



(60.62 * 40.3 *10^-6) / (54.94 *0.100) = 4.45 *10^-4 M

The 10^-6 was used as a conversion factor from micrograms to grams.

I'm not sure if this is right, as it is not taking into account the H3PO4 added.
 
Last edited:
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What is the question?

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Sorry! The question is to find the final molarity of the solution.
 
Molarity of permanganate?

How many moles of permanganate created per mole of manganese?

georgiabrown said:
C Mn2+ = 0.06061 M (60.61 micrograms/ mL)

Are you sure about it? M usually stands just for mol/L.

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Last edited:
Yes, the molarity of permanganate. The concentration on Mn2+ was initially given in ppm (micrograms/mL) and i converted that value to mol/L...
 
If you have converted 60.61 ppm to 0.06061 M you did it wrong.

You have not answered my question.

How many moles of manganese initially?

How many moles of permanganate created per mole of manganese?

What is the final volume?

What is concentration definition?

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methods
 
I think i corrected my conversion : 60.61 ug/mL = 1.10 * 10^-3 M

1. n = cv
= 4.43 *10^-5
2. 1:1 mole ratio
3. final volume = 0.1 L
4. concentration = 4.43*10^-4

is the H3PO4 not involved in the reaction?
 
I got 4.45x10-4M, that's most likely because of some rounding errors and/ordifferent molar masses used - but in general that's the same.

Phosphoric acid... This is a limiting reagent problem. Think about amounts of substances used.

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