To calculate the mass of sodium carbonate required to remove essentially all of the calcium ion from the solution, we need to determine the number of moles of calcium ions present in the solution and then use the balanced chemical equation to find the corresponding amount of sodium carbonate.
Given: Volume of solution = 750 L Concentration of Ca2+ = 43 mg/L
First, let's calculate the number of moles of calcium ions (Ca2+) in the solution: Concentration of Ca2+ = 43 mg/L
To convert from milligrams (mg) to grams (g): 43 mg = 43/1000 g = 0.043 g
Now, we can calculate the number of moles of Ca2+: Molar mass of Ca2+ = 40.08 g/mol (calcium has an atomic mass of 40.08 g/mol) Moles of Ca2+ = mass of Ca2+ / molar mass of Ca2+ = 0.043 g / 40.08 g/mol
Next, we use the balanced chemical equation to determine the stoichiometric ratio between Ca2+ and Na2CO3. According to the equation, 1 mole of Ca2+ reacts with 1 mole of Na2CO3.
From the balanced equation: Ca2+(aq) + CO32-(aq) → CaCO3(s)
Since the stoichiometric ratio is 1:1, the number of moles of Ca2+ will be equal to the number of moles of Na2CO3 required.
Finally, we calculate the mass of sodium carbonate required: Molar mass of Na2CO3 = 105.99 g/mol
Mass of Na2CO3 = moles of Na2CO3 × molar mass of Na2CO3
Substituting the values, we have: Mass of Na2CO3 = 0.043 g × 105.99 g/mol
Therefore, the mass of sodium carbonate required to remove essentially all of the calcium ions from 750 L of the solution is approximately 4.56 grams.