A solution contains 10. mmol of H₃PO₄ and 5.0 mmol of NaH₂PO₄. How many milliliters of 0.10 M NaOH must be added to reach the second equivalence point of the titration of the H₃PO₄ with NaOH? A) 2.0 × 10² mL B) 150 mL C) 250 mL D) 1.0 × 10² mL E) 50 mL

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To reach the second equivalence point, we need to determine how much NaOH is required to react with both the H₃PO₄ and NaH₂PO₄ present in the solution. Since the two compounds have different amounts in the solution, we need to determine which one will be the limiting factor in the reaction. 1. H₃PO₄: 2 moles of NaOH are required to react with 1 mole of H₃PO₄, so for 10 mmol of H₃PO₄, we need 20 mmol of NaOH. 2. NaH₂PO₄: 1 mole of NaOH is required to react with 1 mole of NaH₂PO₄, so for 5 mmol of NaH₂PO₄, we need 5 mmol of NaOH. Since there is more H₃PO₄ present than NaOH, it will be the limiting factor in the reaction. Therefore, we need to add more NaOH to fully react with all the H₃PO₄ present in the solution. To calculate the volume of 0.10 M NaOH needed, we can use the formula: moles of NaOH = Molarity x volume (in liters) For 20 mmol of NaOH, we can solve for the volume of NaOH as follows: 0.020 mol = 0.10 M x volume (in liters) volume (in liters) = 0.20 L Finally, we convert the volume in liters to milliliters: volume (in mL) = 0.20 L x 1000 mL/L = 200 mL Therefore, we need to add 200 mL of 0.10 M NaOH to reach the second equivalence point.

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