Question

Exercises 11.4 12- What volume of 1.000 M Na2CO3 will react with 342 ml of 0.733 M H3PO4? 3Na2CO3 + 2H3PO4 - 2Na3PO4 + 3H2O + 3CO2 Exercise 11.7-end of chapter material For sanitary reasons, water in pools should be chlorinated to a maximum level of 3.0 ppm. In a typical 5,000 gal pool that contains 21,200 kg of water, what mass of chlorine must be added to obtain this concentration?

Answer 1

Answer:

1. As given in the reaction, 3 moles of sodium carbonate requires 2 moles of phosphoric acid for the complete neutralization. (or 1 mole of H₃PO₄ requires 1.5 mL of Na₂CO₃)

Given: 342 mL of 0.733 M H₃PO₄ and 1.0 M of Na₂CO₃

342 mL of solution contains = (0.733/1000) x 342 = 0.2507 moles of H₃PO₄

Hence, moles of Na₂CO₃ requires = 0.2507 x 1.5 = 0.37605 moles

Now molarity of Na₂CO₃ solution = 1.0 M

Hence, 1 mole =1000 mL; So, 0.37605 moles of Na₂CO₃ = (1000/1) x 0.37605 = 376.05 mL of 1.0 M of Na₂CO₃ required

2. For 3.0 ppm concentration

Assuming the density ~1.0g/mL

So the total volume of solution = 21200 kg ~21200 L

Now, 1 ppm is basically = 1 mg/L

Hence, 3ppm = 3mg/L or 3mg of chlorine is present in = 1L of water

So, 21200 L contains = 3 x 21200 = 63600 mg of Chlorine or 63.600 g

Hence, 63.600 g of Chlorine is need to be added to reach this concentration.

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