What mass (grams) of sodium phosphate is required to precipitate all silver ions from 94 mL of a 0.332 M solution of silver nitrate?
To determine the mass of sodium phosphate required to precipitate all the silver ions from the silver nitrate solution, we need to consider the balanced chemical equation for the reaction between sodium phosphate (Na3PO4) and silver nitrate (AgNO3):
3 Na3PO4 + 4 AgNO3 → Ag3PO4 + 3 NaNO3
From the balanced equation, we can see that the ratio of sodium phosphate to silver phosphate is 3:1. This means that three moles of sodium phosphate are required to precipitate one mole of silver phosphate.
Here are the steps to calculate the mass of sodium phosphate:
Step 1: Calculate the number of moles of silver nitrate:
To find the number of moles, we use the formula:
moles = concentration (M) × volume (L)
moles of silver nitrate = 0.332 M × 0.094 L = 0.031208 moles
Step 2: Calculate the number of moles of silver phosphate:
Since the balanced equation indicates that the ratio of sodium phosphate to silver phosphate is 3:1, we can conclude that the number of moles of silver phosphate is the same as the moles of silver nitrate.
moles of silver phosphate = 0.031208 moles
Step 3: Calculate the mass of sodium phosphate:
To find the mass of sodium phosphate, we need to consider the molar mass of sodium phosphate (Mr = 163.94 g/mol).
mass of sodium phosphate = moles of sodium phosphate × molar mass of sodium phosphate
mass of sodium phosphate = 0.031208 moles × 163.94 g/mol
mass of sodium phosphate = 5.11 grams (rounded to two decimal places)
Therefore, approximately 5.11 grams of sodium phosphate is required to precipitate all the silver ions from the given solution.