How to Calculate mEq/L of Ba from BaBr2 Solution?

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SUMMARY

The discussion focuses on calculating the milliequivalents per liter (mEq/L) of barium (Ba) from a barium bromide (BaBr2) solution. The participant dissolved 875.3 mg of BaBr2 in one liter of water and sought to determine the grams of Ba present in the solution. It was established that Ba has a charge of 2+, indicating that one mole of Ba contributes two equivalents. The participant needed to convert the mass of BaBr2 to grams of Ba using the molar mass of Ba (137.3 g/mol) to complete the calculation.

PREREQUISITES
  • Understanding of molar mass and conversions between grams and moles
  • Knowledge of equivalents and their relationship to ionic charge
  • Familiarity with the concept of milliequivalents (mEq)
  • Basic chemistry concepts related to ionic compounds and dissolution
NEXT STEPS
  • Calculate the grams of Ba in BaBr2 using its molar mass (BaBr2: 137.3 g/mol for Ba and 79.9 g/mol for Br)
  • Learn about the concept of equivalents in chemistry and how to calculate them
  • Study the process of converting moles to milliequivalents in ionic solutions
  • Explore the implications of ionic charges on equivalent calculations in various compounds
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Chemistry students, educators, and professionals involved in analytical chemistry or solution preparation, particularly those focusing on ionic compounds and their concentrations.

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Homework Statement

Barium bromide BaBr2 (875.3 mg) was dissolved in water to make one liter of solution. Express the concentration in the following units: mEq/L of Ba.



Homework Equations

Ba has a charge of 2+ which means it has 2 Eq in 1 mole.



The Attempt at a Solution

First I changed the 875.3 mg into .8753 g of BaBr2. The thing that really confuses me is how do I know how many grams of Ba is in .8753 g of BaBr2? I understand how to do this problem if it asked for the mEq/L of BaBr2, but I'm stuck because I can't figure out how many grams of just Ba is in .8753g of BaBr2. Once I figure out the g of Ba I can convert it into moles since there are 137.3 g per mole of Ba.
 
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Why couldn't you do the same thing for BaBr2? Equivalents don't 'equate' to charge either.

example

H3PO4 <-------> 3H+ + PO4-3

In this example, there are three equivalents of H+ per mole of H3PO4 and only one equivalent of PO4-3. Notice how the number of equivalents are unrelated to the charge on the ions generated?
 

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