Question

What volume of \(0.0521 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) 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.

Short answer

To neutralize exactly \(14.20 \ \text{mL}\) of \(0.141 \ \text{M} \ \text{H}_{3}\text{PO}_{4}\), approximately \(0.0256 \ \text{L}\) (or \(25.6 \ \text{mL}\)) of \(0.0521 \ \text{M} \ \text{Ba(OH)}_{2}\) solution is required.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between barium hydroxide and phosphoric acid is: \(2 \mathrm{Ba(OH)_2 + 3H_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6H_2O\)

Determine moles of H3PO4

First, we need to determine the moles of H3PO4 in the given volume. We can do this using the given concentration and volume. The formula is: Moles of H3PO4 = (concentration of H3PO4) × (volume of H3PO4) Moles of H3PO4 = (0.141 mol/L) × (14.20 mL × (1 L / 1000 mL)) Moles of H3PO4 = 0.0020022 mol

Use stoichiometry to determine moles of Ba(OH)2

From the balanced chemical equation, we know that 2 moles of Ba(OH)2 are needed to neutralize 3 moles of H3PO4. Using this stoichiometric ratio, we can determine the moles of Ba(OH)2 required: You have \(\frac{2}{3}\) moles of \(Ba(OH)_2\) per mole of \(H_3PO_4\) Moles of Ba(OH)2 = Moles of H3PO4 × \(\frac{2 \ \text{moles Ba(OH)2}}{3 \ \text{moles H3PO4}}\) Moles of Ba(OH)2 = 0.0020022 mol × \(\frac{2}{3}\) = 0.0013348 mol

Calculate the volume of Ba(OH)2

Now we can use the concentration of Ba(OH)2 to determine the volume required. The formula is: Volume of Ba(OH)2 = \(\frac{\text{Moles of Ba(OH)2}}{\text{Concentration of Ba(OH)2}}\) Volume of Ba(OH)2 = \(\frac{0.0013348 \ \text{mol}}{0.0521 \ \text{mol/L}}\) Volume of Ba(OH)2 ≈ 0.0256 L