Consider the reaction of 14.20ml of 0.141 M H3PO4 and 0.0521 M Ba(OH)2.
2H3PO4 + 3Ba(OH)2 ->Ba3(PO4)2 + 6HOH
What volume of Ba(OH)2 is required to completely react with the H3PO4?
To determine the volume of Ba(OH)2 required to completely react with the H3PO4, we need to use the stoichiometry of the balanced equation.
The balanced equation shows that 2 moles of H3PO4 react with 3 moles of Ba(OH)2. This means that the mole ratio between H3PO4 and Ba(OH)2 is 2:3.
First, let's calculate the moles of H3PO4 in the solution:
Molarity of H3PO4 = 0.141 M
Volume of H3PO4 = 14.20 mL = 0.01420 L
moles of H3PO4 = Molarity x Volume
moles of H3PO4 = 0.141 M x 0.01420 L
Next, let's find the volume of Ba(OH)2 required to react with the H3PO4. Since the mole ratio between H3PO4 and Ba(OH)2 is 2:3, the moles of Ba(OH)2 is related to the moles of H3PO4 by:
moles of Ba(OH)2 = (moles of H3PO4) x (3/2)
Finally, we can calculate the volume of Ba(OH)2 using the molarity and moles of Ba(OH)2:
Molarity of Ba(OH)2 = 0.0521 M
Volume of Ba(OH)2 = ?
moles of Ba(OH)2 = (moles of H3PO4) x (3/2)
Volume of Ba(OH)2 = (moles of Ba(OH)2) / (Molarity of Ba(OH)2)
Putting it all together:
moles of H3PO4 = 0.141 M x 0.01420 L
moles of Ba(OH)2 = (moles of H3PO4) x (3/2)
Volume of Ba(OH)2 = (moles of Ba(OH)2) / (Molarity of Ba(OH)2)
To determine the volume of Ba(OH)2 required to completely react with H3PO4, we need to use the concept of stoichiometry and the given concentrations of H3PO4 and Ba(OH)2.
1. Write down the balanced equation for the reaction:
2H3PO4 + 3Ba(OH)2 → Ba3(PO4)2 + 6H2O
2. Convert the volume of H3PO4 to moles:
Given volume of H3PO4 = 14.20 mL
Since the concentration of H3PO4 is 0.141 M (moles per liter), we can convert the volume of H3PO4 to moles using the formula:
moles = concentration (M) × volume (L)
moles of H3PO4 = 0.141 M × (14.20 mL / 1000 mL/L)
moles of H3PO4 = 0.0020034 mol
3. Use the stoichiometry of the balanced equation to calculate the moles of Ba(OH)2 needed:
From the balanced equation, we can see that 2 moles of H3PO4 react with 3 moles of Ba(OH)2.
Therefore, the moles of Ba(OH)2 needed can be calculated as:
moles of Ba(OH)2 = (3 moles Ba(OH)2 / 2 moles H3PO4) × moles of H3PO4
moles of Ba(OH)2 = (3 / 2) × 0.0020034 mol
moles of Ba(OH)2 = 0.0030051 mol
4. Convert the moles of Ba(OH)2 to volume:
Given concentration of Ba(OH)2 = 0.0521 M
We can use the formula, moles = concentration × volume, to calculate the volume:
volume of Ba(OH)2 = (moles of Ba(OH)2 / concentration of Ba(OH)2)
volume of Ba(OH)2 = 0.0030051 mol / 0.0521 M
volume of Ba(OH)2 = 0.0577 L = 57.7 mL
Therefore, 57.7 mL of Ba(OH)2 is required to completely react with 14.20 mL of 0.141 M H3PO4.