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?

Question

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To reach the second equivalence point, we need to add enough NaOH to react with both the H₃PO₄ and the NaH₂PO₄ in the solution. Since NaHS is a strong acid, it will be completely ionized in solution, so we can consider it as a source of H+ ions. The balanced chemical equation for the reaction between H₃PO₄ and NaOH is: H₃PO₄ + NaOH → Na₂PO₄ + H2O At the first equivalence point, all of the H₃PO₄ has reacted with NaOH, leaving only NaHS and some excess NaOH in solution. At the second equivalence point, all of the NaHS has reacted with NaOH, leaving only water and some excess NaOH in solution. To calculate the amount of NaOH needed to reach the second equivalence point, we need to consider the stoichiometry of the reaction. From the balanced chemical equation, we can see that the molar ratio of H₃PO₄ to NaOH is 1:1. This means that 10 mmol of H₃PO₄ will react with 10 mmol of NaOH. The molar ratio of NaHS to NaOH is also 1:1, so the 5.0 mmol of NaHS will also react with 5.0 mmol of NaOH. Therefore, the total amount of NaOH needed to reach the second equivalence point is: 10 mmol H₃PO₄ + 5.0 mmol NaHS = 15 mmol NaOH To calculate the volume of 0.10 M NaOH needed, we can use the following equation: moles of NaOH = Molarity x Volume (in liters) 15 mmol NaOH = 0.10 M x Volume (in liters) Volume (in liters) = 15 mmol NaOH / (0.10 M) = 0.15 L Finally, we convert the volume to milliliters: Volume (in milliliters) = 0.15 L x 1000 mL/L = 150 mL Therefore, the answer is choice A) 250 mL.

A Solution Contains 10 ×\times× 102 D)50 E)2 ×\times× 102

A Solution Contains 10 $\times$ 10²
D)50
E)2 $\times$ 10²