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?
mols H3PO4 ==> M x L = ?
Use the coefficients in the balanced equation to convert mols H3PO4 to mols Ba(OH)2.
Now M Ba(OH)2 = mols Ba(OH)2/L Ba(OH)2. You know mols and M, calculate L.
To find the volume of Ba(OH)2 required to completely react with the H3PO4, we can use the concept of stoichiometry and the given concentrations and volumes.
First, we need to determine the amount of H3PO4 in moles. To do this, we multiply the concentration of H3PO4 by the volume in liters:
0.141 M * 0.01420 L = 0.002002 moles
According to the balanced chemical equation, the stoichiometric ratio between H3PO4 and Ba(OH)2 is 2:3. Therefore, we can determine the amount of Ba(OH)2 needed by multiplying the amount of H3PO4 by the stoichiometric ratio:
0.002002 moles of H3PO4 * (3 moles of Ba(OH)2 / 2 moles of H3PO4) = 0.003003 moles of Ba(OH)2
Now, we can calculate the volume of Ba(OH)2 solution needed to provide this amount of Ba(OH)2 using its concentration:
0.003003 moles / 0.0521 M = 0.0576 L
Finally, we convert this volume to milliliters:
0.0576 L * 1000 mL/L = 57.6 mL
Therefore, approximately 57.6 mL of Ba(OH)2 solution is required to completely react with 14.20 mL of 0.141 M H3PO4.