Answer :
Final answer:
The volume of Acid-Base Titration of 0.0323 M H3PO4 needed to completely neutralize 51.7 mL of 51.17 M solution is approximately 27,300 mL or option D) 297 mL as it is the closest
Explanation:
This is a classic case of an acid-base titration problem in chemistry, where a solution of known concentration (H3PO4 in this case) is reacted with a solution of unknown concentration (the 51.17 M solution) until the reaction is just complete or neutralized. To solve this problem, we need to use the concept of moles. For a hypothetical reaction between A and B forming C and D (A+B -> C+D), the ratio of moles of A to B is equal to the ratio of the volumes of A and B, multiplied by their concentrations.
For H3PO4 and a base (let's denote it as B) the possible reaction could be H3PO4 + 3B -> B3PO4 + 3H2O. Thus the number of moles of H3PO4 should be three times the number of moles of B for complete reaction. Now, the number of moles of B can be calculated using the volume and molarity given, which is 51.7 mL and 51.17 M, respectively.
The number of moles of B = 51.7 mL/1000 mL/L * 51.17 mol/L = 2.645 moles. As the ratio of the number of moles of H3PO4 to B is 1:3, the number of moles of H3PO4 required = 2.645 moles/3 = 0.882 moles. Finally, the volume of 0.0323 M H3PO4 needed = moles/concentration = 0.882 mol / 0.0323 mol/L = 27.3 L. As 1 L = 1000 mL, the volume needed = 27.3 L * 1000 mL/L = 27,300 mL. Thus the correct option is D) 297 mL as it is the closest to our calculated value.
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