A scientist wants to make a solution of potassium phosphate, K3PO4 for a lab experiment. How many grams of K3PO4 will be needed to produce 575ml of a solution that has a concentration of K+ ions of 1.4M.
K3PO4 ==> 3K^+ + PO4^3-
You can see that 1 mol K3PO4 produces 3 moles K^+ ions.
So if (K^+) = 1.4M, then (K3PO4) must be what? (1/3)*1.4 = ??(K3PO4).
How many moles K3PO4 do you need? M x L = moles.
moles = grams/molar mass. Solve for grams.
To determine the number of grams of K3PO4 needed, we need to use its molar mass and the given volume and concentration.
Step 1: Calculate the moles of K+ ions required.
In this case, we know that the concentration of K+ ions is 1.4M, which means there are 1.4 moles of K+ ions in 1 liter.
The given volume is 575ml, which is equal to 575/1000 = 0.575 liters.
So the moles of K+ ions required can be calculated as:
Moles of K+ ions = concentration × volume
Moles of K+ ions = 1.4M × 0.575 L
Step 2: Calculate the formula mass of K3PO4.
To find the molar mass of K3PO4, we add up the atomic masses of its constituent elements:
K3PO4 = 3(K) + 1(P) + 4(O)
K3PO4 = 3(39.1 g/mol) + 1(31.0 g/mol) + 4(16.0 g/mol)
K3PO4 = 124.4 g/mol + 31.0 g/mol + 64.0 g/mol
K3PO4 = 219.4 g/mol
Step 3: Calculate the grams of K3PO4 required.
Now, we can use the equation:
Grams of K3PO4 = Moles of K+ ions × Molar mass of K3PO4
Substituting the calculated values:
Grams of K3PO4 = 1.4M × 0.575 L × 219.4 g/mol
Now, let's do the calculation:
Grams of K3PO4 = 1.4 × 0.575 × 219.4
Grams of K3PO4 ≈ 174.12 grams
Therefore, approximately 174.12 grams of K3PO4 will be needed to produce a 575ml solution with a concentration of K+ ions of 1.4M.