What volume of 0.0521\(M \mathrm{Ba}(\mathrm{OH})_{2}\) is required to neutralize exactly 14.20 \(\mathrm{mL}\) of 0.141 $\mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4} ?$ Phosphoric acid contains three acidic hydrogens.

To find the moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\), use the formula: moles = molarity × volume Here, the molarity is 0.141 M, and the volume is 14.20 mL (which should be converted to Liters to match the units). moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) = \(0.141\,\text{M} \times 0.01420\,\text{L}\) = 0.0020022 moles

Since phosphoric acid has three acidic hydrogens and barium hydroxide has two hydroxide ions, the balanced neutralization reaction can be written as: \(2 \mathrm{H}_{3} \mathrm{PO}_{4} + 3 \mathrm{Ba}(\mathrm{OH})_{2} \rightarrow 6 \mathrm{H}_{2}\mathrm{O} + 2 \mathrm{Ba}_{3} (\mathrm{PO}_4)_2\) From the stoichiometry of the balanced equation, 2 moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) require 3 moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\). So, moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) can be determined as follows: moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) = \(\frac{3}{2}\times\) moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) = \(\frac{3}{2}\times 0.0020022 = 0.0030033\,\text{moles}\)

Now we know the moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) required, we can use the molarity of the \(\mathrm{Ba}(\mathrm{OH})_{2}\) solution to determine the volume needed to neutralize the \(\mathrm{H}_{3} \mathrm{PO}_{4}\) solution. We will use the formula: volume = \(\frac{\text{moles}}{\text{molarity}}\) volume of \(\mathrm{Ba}(\mathrm{OH})_{2}\) needed = \(\frac{0.0030033\,\text{moles}}{0.0521\,\text{M}}\) = 0.0576 L To convert the volume to milliliters, multiply by 1000: volume of \(\mathrm{Ba}(\mathrm{OH})_{2}\) needed = 57.6 mL So, 57.6 mL of 0.0521 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) is required to neutralize exactly 14.20 mL of 0.141 M \(\mathrm{H}_{3} \mathrm{PO}_{4}\).