A hard water sample contains 121 mg of CaCO3 per liter of water. Calculate the mass of Na3Po4 needed to remove all Ca^2+ ions from 2.50 L of the water sample?
I don't understand what u mean by substitute into ksp expression & solve for PO4^3?
To calculate the mass of Na3PO4 needed to remove all Ca^2+ ions from the water sample, we need to follow the steps below:
Step 1: Convert the given concentration of CaCO3 from mg/L to moles/L.
Given: Hard water sample contains 121 mg of CaCO3 per liter of water.
The molar mass of CaCO3 is 100.09 g/mol.
Conversion: (121 mg/L) / (100.09 g/mol) = 0.1209 mmol/L
Step 2: Calculate the number of moles of Ca^2+ ions present in the water sample.
For every mole of CaCO3, there is one mole of Ca^2+ ions.
So, the number of moles of Ca^2+ ions = 0.1209 mmol/L.
Step 3: Calculate the amount of Na3PO4 needed to remove all Ca^2+ ions.
The balanced equation for the reaction between Ca^2+ and Na3PO4 is:
3Ca^2+(aq) + 2PO4^3-(aq) → Ca3(PO4)2(s)
According to the balanced equation, it takes 3 moles of Ca^2+ ions to react with 2 moles of Na3PO4.
Therefore, the amount of Na3PO4 needed = (0.1209 mmol/L) × (2 mol Na3PO4 / 3 mol Ca^2+).
Step 4: Convert the amount of Na3PO4 needed to mass.
The molar mass of Na3PO4 is 163.94 g/mol.
Conversion: (0.1209 mmol/L) × (2 mol Na3PO4 / 3 mol Ca^2+) × (163.94 g/mol) = mass of Na3PO4 needed in grams.
Now let's calculate the mass of Na3PO4 needed:
(0.1209 mmol/L) × (2 mol Na3PO4 / 3 mol Ca^2+) × (163.94 g/mol) = 0.130 grams of Na3PO4
Therefore, you would need approximately 0.130 grams of Na3PO4 to remove all the Ca^2+ ions from 2.50 L of the hard water sample.
Ca3(PO4)2 ==> 3Ca^2+ + 2PO4^3-
Ksp Ca3(PO4)2 = ?. Look that up in your text.
Ksp = (Ca^2+)^3(PO4^3-)^2
(Ca^2+) = 0.121/molar mass CaCO3.
Substitute into Ksp expression, solve for (PO4^3-) (note this will be in mols/L), multiply by 2.5 to convert to 2.5 L, then grams Na3PO4 = mols PO4^3- x molar mass Na3PO4.