What mass of KH2PO4 is needed to prepare 500.0 mL of a 0.067M solution?
How many moles do you need? That is M x L = moles.
Then moles = grams/molar mass. Solve for grams.
To find the mass of KH2PO4 needed to prepare a certain volume of a solution, we need to use the formula:
mass (g) = molarity (M) × volume (L) × molar mass (g/mol)
First, let's find the molar mass of KH2PO4.
The molar mass of potassium (K) is about 39.10 g/mol.
The molar mass of hydrogen (H) is about 1.01 g/mol.
The molar mass of phosphorus (P) is about 30.97 g/mol.
The molar mass of oxygen (O) is about 16.00 g/mol (there are four oxygen atoms in KH2PO4).
Calculating the molar mass of KH2PO4:
molar mass (KH2PO4) = (molar mass of K) + 2 × (molar mass of H) + (molar mass of P) + 4 × (molar mass of O)
= 39.10 + 2 × 1.01 + 30.97 + 4 × 16.00
= 136.09 g/mol
Now, we can use the formula to calculate the mass of KH2PO4 needed:
mass (g) = 0.067 M × 0.500 L × 136.09 g/mol
= 4.586 g
Therefore, approximately 4.586 grams of KH2PO4 is needed to prepare 500.0 mL of a 0.067 M solution.