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Calculate the milli-equivalents of washing soda required to remove hardness from one liter of hard water containing 18.00 mg of Mg²+.

a) 0.9 mEq
b) 1.8 mEq
c) 3.6 mEq
d) 5.4 mEq

Answer :

Final answer:

To find the milli-equivalents of washing soda required to remove Mg²+ from hard water, one must calculate the moles of Mg²+ present, establish the equivalent weight of washing soda, and convert moles into milli-equivalents. The calculation concludes with a requirement of 0.740 mEq of washing soda to remove the hardness from one liter of water containing 18.00 mg of Mg²+.

Explanation:

The question 'Calculate the milli-equivalents of washing soda required to remove hardness from one liter of hard water containing 18.00 mg of Mg²+' pertains to a stoichiometric conversion from the mass of Mg²+ to milli-equivalents (mEq) of Na2CO3, commonly known as washing soda.

First, calculate the moles of Mg²+ in 18.00 mg by using the molar mass of Mg (24.305 g/mol). This gives us 18.00 mg × (1 g / 1000 mg) / 24.305 g/mol = 0.000740 mol Mg²+.

Since washing soda has a molar mass of 105.99 g/mol and the equivalent weight is half the molar mass (due to the formation of two equivalents per mole Na2CO3), we find the equivalent weight to be 52.995. The equivalent weight tells us how much of a substance is needed to react with or supply one mole of hydrogen ions (H+) or hydroxyl ions (OH-), or to react with or supply one mole of divalent ions such as Mg²+.

Each divalent Mg ion will need one equivalent of Na2CO3. Therefore, 0.000740 mol of Mg²+ requires 0.000740 equivalents of Na2CO3. To find the milli-equivalents, we multiply by 1000, resulting in 0.740 mEq of Na2CO3. Thus, the answer is slightly less than the given option a) 0.9 mEq.

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