What volume of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid according to

Chemistry

1 answer:

kirill115 [55]2 years ago

5 0

Taking into account the definition of molarity and the stoichiometry of the reaction, the correct option is option C) 0.44 L of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid.

The balanced reaction is:

H₃PO₄ (aq) + 3 NaOH (aq) → Na₃PO₄ (aq) + 3 H₂O(aq)

Then, by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

H₃PO₄: 1 mole
NaOH: 3 moles
Na₃PO₄: 1 mole
H₂O: 3 moles

Molarity is the number of moles of solute that are dissolved in a given volume.

Molarity is determined by:

$Molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

In this case, 0.42 L of 2.39 M phosphoric acid reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:

$2.39 \frac{moles}{liter}=\frac{number of moles of phosphiric acid}{0.42 liters}$

Solving:

number of moles of phosphiric acid= 2.39 $\frac{moles}{liter}$ * 0.42 liters

number of moles of phosphiric acid= 1.0038 moles ≅ 1 mole

Approaching 1 mole of the amount of phosphoric acid required, then by stoichiometry of the reaction, 3 moles of NaOH are necessary to react with 1 mole of the acid.

Then by definition of molarity and knowing that 6.9 M NaOH is needed, you can calculate the necessary volume amount of NaOH by:

$6.9 \frac{moles}{liter} =\frac{3 moles}{volume}$