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)250 mL B)2.0 × 10² mL C)50 mL D)150 mL E)1.0 × 10² mL

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To reach the second equivalence point in the titration of $H_2PO_4^-$ with NaOH, all $H_2PO_4^-$ must be converted to $PO_4^{3-}$. The solution initially contains 10 mmol of $H_2PO_4^-$ and 5 mmol of $NaHPO_4^-$, which does not require titration to reach the second equivalence point. To convert 10 mmol of $H_2PO_4^-$ to $PO_4^{3-}$, 10 mmol of NaOH is needed. Since the NaOH solution is 0.10 M, the volume of NaOH required is $V = \frac{n}{C} = \frac{10\,mmol}{0.10\,M} = 100\,mL$. However, to reach the second equivalence point, an additional 5 mmol of NaOH is needed to convert $NaHPO_4^-$ to $PO_4^{3-}$, totaling 15 mmol of NaOH. Thus, the total volume of NaOH required is $V = \frac{15\,mmol}{0.10\,M} = 150\,mL$.

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