Did I Mix the Solutions Correctly?

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SUMMARY

The discussion revolves around two chemistry problems involving the preparation of solutions and concentration calculations. For the first problem, the user calculated that 5.56ml of concentrated (18M) sulfuric acid is needed to prepare 50ml of 2M solution, while the book suggests 16ml. The second problem involves calculating the new concentration of a toxic compound in a pond after the addition of liquid waste. The user arrived at a concentration of 0.42ppm, while the book states 0.49ppm. The key error identified was the need to use volume/volume (v/v) concentration instead of mass/volume (g/v) for the second problem.

PREREQUISITES
  • Understanding of molarity and solution preparation
  • Knowledge of parts per million (ppm) calculations
  • Familiarity with volume/volume (v/v) concentration concepts
  • Basic skills in unit conversions and dimensional analysis
NEXT STEPS
  • Review the principles of molarity and how to prepare solutions accurately
  • Study the calculations for parts per million (ppm) and its applications in environmental chemistry
  • Learn about volume/volume (v/v) concentration and its significance in solution chemistry
  • Practice unit conversion techniques, especially in the context of chemical solutions
USEFUL FOR

Chemistry students, environmental scientists, and professionals involved in chemical solution preparation and concentration calculations will benefit from this discussion.

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Homework Statement


Here is two similar questions but which I got a different answer to the back of the book

1. You are to prepare 50ml of 2M sulfuric acid, which is to be added to a sample of white wine to make the sulfur dioxide preservative in the wine available in solution, so that its concentration can be determined. What volume of concentrated (18M) acid will you prepare?

2. A factory discharged 15ML of a liquid waste containing 12ppm (v/v) of a highly toxix compound into a pond. Before it was added, this pond already contained 920ML of wastewater in which this compound was present at a concentration of 0.23ppm (v/v). What was the new concentration of this compound in this pond?


Homework Equations


conversions


The Attempt at a Solution


Q1.
2 moles/litre, 50ml => 0.1moles

18moles/litre. Need 0.1moles so 0.1/18=0.0056L=5.56ml but answers suggest 16ml


Q2.
ppm (v/v) => 1g/10^6g = 1g/1000kg

1L=1kg assuming solvent is water

15ML @ 12ppm => 15*10^6 kg @ 12g/1000kg => 15*10^3 * 12g grams of solute or 180kg of solute

920ML @ 0.23ppm => 920*10^6 kg @ 0.23g/1000kg => 920*10^3 * 0.23g or 211.6kg of solute

Hence total solute is 391.6kg. Total solution is 935ML. So a new total concentration of 391.6kg/(935*10^6kg) * 10^6g = 0.42g/10^6g or 0.42ppm. But the answers suggest 0.49ppm.

Who is wrong?
 
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1. You are OK.

2. You are OK and wrong at the same time. Your result is correct numerically, but note that question is about v/v concetration - so you should calculate using L, not g.


concentration lectures
 
v/v => ml of solute per 100ml of solution. Luck in solution of water, 1ml = 1g. I used 1g in my calculations. So all I had to do was to switch g to ml. The magnitude of the numbers wouldn't change. Thanks for picking that up for me. I will be more careful next time.
 

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