What is the \(\mathrm{pH}\) of a solution made by mixing \(0.40 \mathrm{~mol}\) \(\mathrm{NaOH}, 0.25 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{HPO}_{4}\), and \(0.30 \mathrm{~mol} \mathrm{H}_{3} \mathrm{PO}_{4}\) with water and diluting to \(2.00 \mathrm{~L} ?\)

First, identify the chemical reactions that will occur when NaOH reacts with H₃PO₄ and Na₂HPO₄. These reactions will be: \[ \begin{cases} \mathrm{H_{3}PO_{4} + NaOH\ \rightarrow H_{2}PO_{4}^{-} + Na^{+} + H_{2}O} \\ \\ \mathrm{Na_{2}HPO_{4} + NaOH\ \rightarrow Na_{3}PO_{4} + H_{2}O} \end{cases} \]

In order to verify which reaction (if not both) will occur, we need to find the limiting reactant. Calculate the moles of NaOH needed to react with H₃PO₄ and Na₂HPO₄. Moles of NaOH required to react with \(0.30 \mathrm{~mol}\) \(H_{3}PO_{4}\): \(0.30 \cdot {1}/{1} = 0.30 \mathrm{~mol}\), Moles of NaOH required to react with \(0.25 \mathrm{~mol}\) \(Na_{2}HPO_{4}\): \(0.25 \cdot {1}/{1} = 0.25 \mathrm{~mol}\), Total moles of NaOH required: \(0.30 + 0.25 = 0.55 \mathrm{~mol}\) Since we have \( 0.40 \mathrm{~mol}\) of NaOH initially, it is the limiting reactant.

Calculate the molar concentrations of each reactant after the reaction has occurred, using the initial moles of reactants and the limiting reactant (NaOH). Moles of NaOH left: \(0 \mathrm{~mol}\), Moles of H₃PO₄ left: \(0.30 - 0.40 = -0.10 \mathrm{~mol} \to 0 \mathrm{~mol}\) (since it reacts completely), Moles of Na₂HPO₄ left: \(0.25 - (0.40 - 0.30) = 0.15 \mathrm{~mol}\), Moles of \(H_{2}PO_{4}^{-}\) formed: \(0.40 - (0.40 - 0.30) = 0.30 \mathrm{~mol}\), Now, divide the moles by the total volume of the solution to get molar concentrations: \[[H_{3}PO_{4}] = \frac{0 \mathrm{~mol}}{2.00 \mathrm{~L}} = 0 \mathrm{M},\] \[[Na_{2}HPO_{4}] = \frac{0.15 \mathrm{~mol}}{2.00 \mathrm{~L}} = 0.075 \mathrm{M},\] \[[H_{2}PO_{4}^{-}] = \frac{0.30 \mathrm{~mol}}{2.00 \mathrm{~L}} = 0.15 \mathrm{M}.\]

Since we have a mixture of \(H_{2}PO_{4}^{-}\) and \(Na_{2}HPO_{4}\), we can consider the mixture as a buffer solution. Use the Henderson-Hasselbalch equation to find the pH of the solution: \[pH = pK_{a2} + \log{\frac{[A^{-}]}{[HA]}}\] Where \(pK_{a2}\) is the second ionization constant of phosphoric acid (\(H_{3}PO_{4}\)). From a table of ionization constants, we find that \(pK_{a2} \approx 7.21\). Now, plug in the values for the concentrations of \(H_{2}PO_{4}^{-}\) and \(Na_{2}HPO_{4}\): \[pH = 7.21 + \log{\frac{0.075}{0.15}}\] \[pH \approx 6.92\] So, the pH of the solution made by mixing the given reagents and diluting to \(2.00 \mathrm{~L}\) is approximately \(6.92\).