What mass of lithium phosphate is needed to prepare 500.0 mL of a solution having a lithium ion concentration of 0.1385 M?
Li3PO4.
You want (Li^+) = 0.1385M
You want to make this out of Li3PO4. But since Li3PO4 furnishes 3 Li ions for every molecule of Li3PO4, you only need 1/3 that number or 0.1385/3 = ? mols Li3PO4.
Then g Li3PO4 = mols x molar mass.
I did that but got 5.346 g as my answer. Is that correct?
No and partly because I may have led you astray. The numbers I gave you are for 1L of the solution at that concn. You only want 500 mL of that (which I missed on first reading) so you need only 1/2 of your final figure or 5.346/2 = ?.
To determine the mass of lithium phosphate needed, we first need to understand the relationship between concentration, volume, and molar mass.
The formula we will use is:
C = n/V
where C is the concentration in moles per liter (M), n is the number of moles, and V is the volume in liters.
To calculate the number of moles (n), we can use the formula:
n = C x V
where C is the concentration in moles per liter (M), and V is the volume in liters.
In this case, the lithium ion concentration is given as 0.1385 M and the volume is 500.0 mL, which needs to be converted to liters. To convert milliliters to liters, we use the conversion factor:
1 L = 1000 mL
Therefore, 500.0 mL is equal to 500.0 mL/1000 mL/L = 0.5000 L.
Now, we can calculate the number of moles (n):
n = 0.1385 M x 0.5000 L
n = 0.06925 moles
To determine the mass of lithium phosphate (Li3PO4), we need to know its molar mass. The molar mass of lithium phosphate is calculated by adding up the atomic masses of its constituent elements: 3 lithium (Li) atoms and 1 phosphorus (P) atom.
The atomic mass of lithium (Li) is approximately 6.941 g/mol, and the atomic mass of phosphorus (P) is approximately 30.97 g/mol.
Therefore, the molar mass of lithium phosphate (Li3PO4) is:
(3 x atomic mass of Li) + atomic mass of P + (4 x atomic mass of O)
= (3 x 6.941 g/mol) + 30.97 g/mol + (4 x 16.00 g/mol)
= 109.87 g/mol
Finally, we can calculate the mass of lithium phosphate (Li3PO4) needed using the following formula:
Mass = n x Molar mass
Mass = 0.06925 moles x 109.87 g/mol
Mass = 7.58 grams (rounded to two decimal places)
Therefore, approximately 7.58 grams of lithium phosphate is needed to prepare a 500.0 mL solution with a lithium ion concentration of 0.1385 M.