Calculate mL (4 sf) of 0.7500 M sodium hydroxide required to neutralize 35.00 mL of 0.7500 M phosphoric acid. Please input number as whole number not scientific notation

105mL of 0.7500M of NaOH are required

Explanation:

Phosphoric acid, H₃PO₄, reacts with sodium hydroxide, NaOH, as follows:

H₃PO₄ + 3 NaOH → Na₃PO₄ + 3H₂O

Where 1 mole of phosphoric acid reacts with 3 moles of NaOH

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:

Moles H3PO4:

35.00mL = 0.03500L * (0.7500mol / L) = 0.2625 moles H3PO4

Moles NaOH:

0.2625 moles H3PO4 * (3mol NaOH / 1mol H3PO4) = 0.07875 moles NaOH

Volume NaOH:

0.07875 moles NaOH * (1L / 0.7500mol) = 0.105L =

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.