A stock solution is prepared by adding 20mL of 0.2M Na3PO4 to enough water to make 80mL. What is the Na+ concentration of 20mL of the stock solution?
I am thinking .012:
a) .02 times .2 = .004
b).004 times 3 = .012 (for each Na in the molecule)
Seems too simple
Yes and no. Yes, it's fairly simple; no 0.012 isn't right.
How much is the Na3PO4 diluted? It started at 0.2M so it is now
0.2M x (20 mL/80 mL) = ?M Na3PO4.
Then Na^+ is 3x that.
Write three equations (including ΔH values) that represent the heats of formation for each C6H12O6(s),
CO2(g), and H2O(l). (Hint: look up heats of formation for each participant)
To find the Na+ concentration in the 20mL stock solution, you need to consider the chemical equation and stoichiometry of the Na3PO4 compound.
The formula of Na3PO4 suggests that there are three Na+ ions for every one Na3PO4 molecule.
First, calculate the total amount of Na3PO4 in the 20mL stock solution:
0.2M represents 0.2 moles of Na3PO4 in 1L of solution.
Therefore, in 20mL (0.02L) of the stock solution, the amount of Na3PO4 can be calculated using the formula:
moles = concentration × volume
moles = 0.2M × 0.02L = 0.004 moles
Next, determine the number of moles of Na+ ions. Since there are three Na+ ions for each Na3PO4 molecule, multiply the moles of Na3PO4 by three:
moles of Na+ ions = 0.004 moles × 3 = 0.012 moles
Finally, calculate the Na+ concentration in the 20mL stock solution:
To convert moles to concentration, divide the moles by the volume:
Na+ concentration = moles of Na+ ions / volume of stock solution
Na+ concentration = 0.012 moles / 0.02L = 0.6M
Therefore, the Na+ concentration in the 20mL stock solution is 0.6M.
Your calculations were correct, and it is indeed that simple! Good job!