A student mixes \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(86.4 \mathrm{~mL}\) of \(0.494 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} .\) Calculate the mass of \(\mathrm{BaSO}_{4}\) formed and the \(\mathrm{pH}\) of the mixed solution.

Based on the exercise, the chemical reaction is: \(Ba(OH)_{2} + H_{2}SO_{4} → BaSO_{4} + 2H_{2}O\). It's balanced because it has equal numbers of each element on both sides. Now, calculate the number of moles (n) of both reactant using their concentrations (C) and volumes (V - in liters), applying the formula \(n = C \times V\):To find the quantity in moles of \(Ba(OH)_{2}\): \(n = 1.00 mol/L * 0.050 L = 0.050 mol\);To find the quantity in moles of \(H_{2}SO_{4}\): \(n = 0.494 mol/L * 0.0864 L = 0.0427 mol\).

The limiting reagent is the one that is completely consumed when the chemical reaction is complete. Since the molar ratio between \(Ba(OH)_{2}\) and \(H_{2}SO_{4}\) is 1:1, \(H_{2}SO_{4}\) is the limiting reagent as it has fewer moles. Use the number of moles of the limiting reagent, \(H_{2}SO_{4}\), to find the moles of the product, \(BaSO_{4}\), using the stoichiometry from the balanced chemical equation. The number of moles of \(BaSO_{4}\) = Moles of \(H_{2}SO_{4}\) = 0.0427 mol.To find the mass (m) of \(BaSO_{4}\) formed, use the formula \(m=nM\), where \(M\) is the molar mass of the compound. The molar mass of \(BaSO_{4}\) is approximately 233.43 g/mol, therefore, \(m = 0.0427 mol*233.43 g/mol = 9.97 g\). So, the mass of \(BaSO_{4}\) formed is 9.97 g.

To calculate the pH, we need to know the concentration of \([OH^{-}]\) ions left in excess from the reactant \(Ba(OH)_{2}\). The remaining moles of \(Ba(OH)_{2}\) are \(0.050 mol - 0.0427 mol = 0.0073 mol\). But, since \(Ba(OH)_{2}\) provides two \([OH^{-}]\) for every mole, the remaining moles of \([OH^{-}]\) are \(0.0073 * 2 = 0.0146 mol\). Now, calculate the total volume of the solution (V_total) to find the concentration of \([OH^{-}]\): \(V_{total} = 0.050 L + 0.0864 L = 0.1364 L\).The concentration of \([OH^{-}]\) is \(0.0146 mol / 0.1364 L = 0.107 mol/L\).To find the pH, first calculate the pOH by applying \(-log[OH^-]\) which gives 0.97. Then, use the relation \(pH + pOH = 14\) at 25°C to find the pH. Finally, pH = 14 - pOH = 14 - 0.97 = 13.03.