The concentration of Ca2+ in a particular water supply is 5.7×10−3M. The concentration of bicarbonate ion, HCO−3, in the same water is 1.7×10−3M. What masses of Ca(OH)2 and Na2CO3 must be added to 5.6×107L of this water to reduce the level of Ca2+ to 23% of its original level?
If you could show the work please that would be fantastic!!!
To solve this problem, we need to calculate the masses of Ca(OH)2 and Na2CO3 needed to reduce the concentration of Ca2+ in the water supply.
First, let's determine the molar masses of Ca(OH)2 and Na2CO3:
- The molar mass of Ca(OH)2:
- Ca: 1 atom x 40.08 g/mol = 40.08 g/mol
- O: 2 atoms x 16.00 g/mol = 32.00 g/mol
- H: 2 atoms x 1.01 g/mol = 2.02 g/mol
- Total molar mass = 40.08 g/mol + 32.00 g/mol + 2.02 g/mol = 74.10 g/mol
- The molar mass of Na2CO3:
- Na: 2 atoms x 22.99 g/mol = 45.98 g/mol
- C: 1 atom x 12.01 g/mol = 12.01 g/mol
- O: 3 atoms x 16.00 g/mol = 48.00 g/mol
- Total molar mass = 45.98 g/mol + 12.01 g/mol + 48.00 g/mol = 106.99 g/mol
Next, we need to calculate the moles of Ca2+ initially present in the water supply and the moles of Ca2+ desired after reducing its concentration.
Moles of Ca2+ initially present:
- Concentration of Ca2+: 5.7 × 10^(-3) M
- Volume of water supply: 5.6 × 10^7 L
- Moles = Concentration × Volume = 5.7 × 10^(-3) M × 5.6 × 10^7 L = 3.192 × 10^5 mol
Moles of Ca2+ desired after reduction:
- Initial moles of Ca2+ × Desired concentration = 3.192 × 10^5 mol × 0.23 = 7.348 × 10^4 mol
Now, let's determine the moles of Ca(OH)2 needed to react with Ca2+ ions:
- Balanced chemical equation between Ca(OH)2 and Ca2+:
Ca(OH)2 + Ca2+ → 2 CaOH+
From the equation, 1 mole of Ca(OH)2 reacts with 1 mole of Ca2+.
Therefore, moles of Ca(OH)2 = moles of Ca2+ desired = 7.348 × 10^4 mol.
To find the mass of Ca(OH)2 needed, we use the formula:
Mass = Moles × Molar mass = 7.348 × 10^4 mol × 74.10 g/mol ≈ 5.444 × 10^6 g
Next, let's determine the moles of Na2CO3 needed to react with bicarbonate ions (HCO3-):
- Balanced chemical equation between Na2CO3 and 2 HCO3-:
Na2CO3 + 2 HCO3- → 2 NaHCO3
From the equation, 1 mole of Na2CO3 reacts with 2 moles of HCO3-.
Therefore, moles of Na2CO3 = 2 × moles of HCO3- = 2 × 1.7 × 10^(-3) M × 5.6 × 10^7 L = 1.904 × 10^5 mol
To find the mass of Na2CO3 needed, we use the formula:
Mass = Moles × Molar mass = 1.904 × 10^5 mol × 106.99 g/mol ≈ 2.036 × 10^7 g
In summary, to reduce the level of Ca2+ to 23% of its original level in the water supply, approximately 5.444 × 10^6 g of Ca(OH)2 and 2.036 × 10^7 g of Na2CO3 must be added to 5.6 × 10^7 L of water.