Answer:
105mL of 0.7500M of NaOH are required
Explanation:
Phosphoric acid, H₃PO₄, reacts with sodium hydroxide, NaOH, as follows:
H₃PO₄ + 3 NaOH → Na₃PO₄ + 3H₂O
<em>Where 1 mole of phosphoric acid reacts with 3 moles of NaOH</em>
<em />
To solve this question we must find, as first, the moles of H3PO4 that react. Using the chemical equation we can find the moles of NaOH required to neutralize this acid. And, with its concentration, we can find the volume oof NaOH required:
<em>Moles H3PO4:</em>
35.00mL = 0.03500L * (0.7500mol / L) = 0.2625 moles H3PO4
<em>Moles NaOH:</em>
0.2625 moles H3PO4 * (3mol NaOH / 1mol H3PO4) = 0.07875 moles NaOH
<em>Volume NaOH:</em>
0.07875 moles NaOH * (1L / 0.7500mol) = 0.105L =
<h3>105mL of 0.7500M of NaOH are required</h3>
