Chemistry problem with molarity and ions

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SUMMARY

The discussion focuses on calculating the percentage by mass of sodium carbonate required in a laundry detergent to precipitate calcium (Ca2+) and magnesium (Mg2+) ions from hard water. The water contains 3.8×10−3 M Ca2+ and 1.1×10−3 M Mg2+, with an average washing machine capacity of 19.0 gallons. The incorrect attempts at solving the problem highlight the need to correctly calculate the moles of each ion and the corresponding mass of sodium carbonate needed for precipitation, rather than simply adding the masses of the ions to the detergent mass.

PREREQUISITES
  • Understanding of molarity and its calculation
  • Knowledge of molar mass (MM) for calcium and magnesium
  • Familiarity with stoichiometry in chemical reactions
  • Basic unit conversions (gallons to liters, kg to grams)
NEXT STEPS
  • Learn how to calculate moles from molarity and volume
  • Study the precipitation reactions involving sodium carbonate and calcium/magnesium ions
  • Review stoichiometric calculations for determining reactant masses
  • Explore the concept of solubility product and its implications in precipitation reactions
USEFUL FOR

Chemistry students, educators, and anyone involved in formulating laundry detergents or studying water chemistry will benefit from this discussion.

ahhppull
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Homework Statement



Sodium carbonate is often added to laundry detergents to soften hard water and make the detergent more effective. Suppose that a particular detergent mixture is designed to soften hard water that is 3.8×10^−3 M in Ca2+ and 1.1×10^−3 M in Mg2+ and that the average capacity of a washing machine is 19.0 gallons of water.

If the detergent requires using 0.63 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.

Homework Equations



Molarity = moles/L
MM Ca = 40.08g
MM Mg = 24.31g
1 kg = .001 g
1 gal = 3.785L

The Attempt at a Solution



I was confused about this problem, but I used 2 of my tries and have 1 try left over. My first answer was 46& and my second answer was 98% (both of these are wrong)

My method's probably wrong, but here's what I did. I started by multiplying 19 gallons by 3.785 to get 71.915 liters. Then I multiplied the molarity of Ca and Mg each by 71.915 liters to get the grams of each. Then I multiplied the moles by the MM of each of the element. Then I added these two up and added it to .63Kkg (630g) and dived 630g by the total mass.
 
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ahhppull said:
My method's probably wrong, but here's what I did. I started by multiplying 19 gallons by 3.785 to get 71.915 liters.

No idea if conversion factor is correct, but the idea is OK.

Then I multiplied the molarity of Ca and Mg each by 71.915 liters to get the grams of each.

I guess you mean moles, not grams.

Then I multiplied the moles by the MM of each of the element. Then I added these two up and added it to .63Kkg (630g) and dived 630g by the total mass.

You calculated total mass of calcium and magnesium dissolved. That's not what you are looking for - you need to add enough sodium carbonate to precipitate both metals.

I guess you are to ignore solubility product and assume reaction goes simply to completion. What reaction? That's what you need to solve the question.

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