Cola contains phosphoric acid, H3PO4. In order to determine the concentration of phosphoric acid in 50mL of Cola, a precipitation reaction with excess silver nitrate was carried out. The product, silver phosphate, was dried and found to weigh 0.0576g. What is the concentration of phosphoric acid in the Cola.
mass Ag3PO4 = 0.0576 grams.
moles Ag3PO4 = 0.0576 g/molar mass Ag3PO4.
mols phosphate = ??
Moles H3PO4 = ??
grams H3PO4 = ??
How do you want concentration calculated. g/mL is g/50 mL.
M = moles/L.
To determine the concentration of phosphoric acid in the Cola, we can use stoichiometry and the mass of the precipitate formed in the precipitation reaction.
First, let's set up the balanced equation for the reaction between phosphoric acid (H3PO4) and silver nitrate (AgNO3):
3AgNO3 + H3PO4 → Ag3PO4 + 3HNO3
From the equation, we can see that 3 moles of silver nitrate react with 1 mole of phosphoric acid to produce 1 mole of silver phosphate.
Next, we need to find the number of moles of silver phosphate using the molar mass of Ag3PO4. The molar mass of Ag3PO4 can be calculated by adding up the atomic masses of the elements:
(3 * atomic mass of Ag) + atomic mass of P + (4 * atomic mass of O)
= (3 * 107.87 g/mol) + 30.97 g/mol + (4 * 16.00 g/mol)
= 411.6 g/mol
Now, we can calculate the number of moles of Ag3PO4 using the mass given:
mass of Ag3PO4 = 0.0576 g
moles of Ag3PO4 = mass of Ag3PO4 / molar mass of Ag3PO4
= 0.0576 g / 411.6 g/mol
≈ 0.00014 mol
Since the stoichiometry of the reaction tells us that 1 mole of Ag3PO4 corresponds to 1 mole of H3PO4, the number of moles of phosphoric acid in the 50 mL of Cola is also 0.00014 mol.
We can now calculate the concentration of phosphoric acid in the Cola:
concentration (in mol/L) = moles of H3PO4 / volume of solution (in L)
The volume of the Cola is given as 50 mL, which is equivalent to 0.0500 L:
concentration = 0.00014 mol / 0.0500 L
= 0.0028 mol/L
Therefore, the concentration of phosphoric acid in the Cola is approximately 0.0028 mol/L.