What volume of 0.0495 M Ba(OH)2 is required to neutralize exactly 14.87 mL of 0.136 M H3PO4? Phosphoric acid contains three acidic hydrogens.
To find the volume of Ba(OH)2 required to neutralize the given amount of H3PO4, we need to determine the stoichiometry of the reaction. The balanced chemical equation for the reaction between Ba(OH)2 and H3PO4 is:
2 Ba(OH)2 + H3PO4 → Ba3(PO4)2 + 2 H2O
From the equation, we can see that 2 moles of Ba(OH)2 react with 1 mole of H3PO4.
Given the concentration and volume of H3PO4, we can calculate the number of moles of H3PO4 using the formula:
moles = concentration × volume
moles of H3PO4 = 0.136 M × 14.87 mL = 2.02032 × 10^-3 moles
Since 1 mole of H3PO4 reacts with 2 moles of Ba(OH)2, we can continue and determine the number of moles of Ba(OH)2 required:
moles of Ba(OH)2 = (2.02032 × 10^-3 moles H3PO4) / (1 mole H3PO4/2 moles Ba(OH)2) = 4.04064 × 10^-3 moles
Finally, we can calculate the volume of Ba(OH)2 using its concentration:
volume of Ba(OH)2 = moles / concentration
volume of Ba(OH)2 = (4.04064 × 10^-3 moles) / (0.0495 M) = 0.0816 L or 81.6 mL
Therefore, 81.6 mL of 0.0495 M Ba(OH)2 is required to neutralize 14.87 mL of 0.136 M H3PO4.