A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 400 mL of a solution that has a concentration of Na+ ions of 1.00 M?
You want how many moles Of Na^+? That is M x L = 1.00 x 0.400 L = 0.400 moles.
How many moles of Na3PO4 will that be? Since there are 3 Na^+ to 1 molecule Na3PO4, 1/3 x 0.400 will be the moles Na3PO4 needed = ??
moles Na3PO4 = grams Na3PO3/molar mass Na3PO4.
Solve for grams.
To calculate the number of grams of Na3PO4 needed, you can use the formula:
moles = concentration (M) × volume (L)
First, convert the volume of the solution from milliliters to liters:
400 mL ÷ 1000 mL/L = 0.4 L
Next, the concentration given is 1.00 M, which means there is 1 mole of Na+ ions per liter of solution. Since Na3PO4 contains 3 Na+ ions per formula unit, we will need to calculate the number of moles of Na3PO4 needed:
moles = concentration (M) × volume (L) = 1.00 M × 0.4 L = 0.4 moles
Lastly, we can calculate the molar mass of Na3PO4 and convert the moles to grams using the following steps:
Molar mass of Na3PO4 = (3 × atomic mass of Na) + atomic mass of P + (4 × atomic mass of O)
= (3 × 22.99 g/mol) + 30.97 g/mol + (4 × 16.00 g/mol)
= 69.00 g/mol + 30.97 g/mol + 64.00 g/mol
= 164.97 g/mol
grams = moles × molar mass = 0.4 moles × 164.97 g/mol = 65.988 g
Therefore, you will need approximately 65.988 grams of Na3PO4 to produce a 400 mL solution with a concentration of Na+ ions of 1.00 M.
To determine the number of grams of tribasic sodium phosphate (Na3PO4) needed to produce a 400 mL solution with a concentration of Na+ ions of 1.00 M, you would need to use the molar mass of Na3PO4, as well as the relationship between molarity, volume, and moles.
Here's how you can calculate it:
Step 1: Find the molar mass of Na3PO4.
The molar mass of Na3PO4 can be calculated by adding up the atomic masses of its constituent elements. The atomic masses of Na, P, and O are 22.99 g/mol, 30.97 g/mol, and 16.00 g/mol, respectively.
The molar mass of Na3PO4 is:
3(Na) + 1(P) + 4(O) = 3(22.99 g/mol) + 30.97 g/mol + 4(16.00 g/mol) = 163.94 g/mol.
Step 2: Use the molarity formula to calculate the moles of Na+ ions.
The molarity (M) is defined as moles of solute divided by the volume of the solution in liters.
Given that the concentration of Na+ ions is 1.00 M and the volume of the solution is 400 mL (which is 0.4 L):
Molarity (M) = moles of solute (Na+ ions) / volume of solution (L)
Rearranging the formula to solve for moles of solute:
moles of solute = Molarity (M) x volume of solution (L)
moles of solute = 1.00 mol/L x 0.4 L = 0.4 mol Na+ ions.
Step 3: Use the stoichiometry of Na3PO4 to determine the moles of Na3PO4 required.
From the balanced chemical equation, you can see that 1 mole of Na3PO4 contains 3 moles of Na+ ions.
Therefore, the moles of Na3PO4 required can be calculated by dividing the moles of Na+ ions by 3:
moles of Na3PO4 = moles of Na+ ions / 3
moles of Na3PO4 = 0.4 mol / 3 = 0.1333... mol Na3PO4.
Step 4: Calculate the mass of Na3PO4 needed.
Finally, to find the mass of Na3PO4 needed, multiply the moles of Na3PO4 by its molar mass:
mass (g) = moles of Na3PO4 x molar mass of Na3PO4
mass (g) = 0.1333... mol x 163.94 g/mol = 21.9 g (rounded to the appropriate number of significant figures).
Therefore, approximately 21.9 grams of Na3PO4 will be needed to produce a 400 mL solution with a concentration of Na+ ions of 1.00 M.