223k views
4 votes
How many mL of a 2.75 M solution of nitric acid is required to neutralize 486.8 mL of 2.94 MNaOH?

User Bluszcz
by
4.8k points

1 Answer

6 votes

Step 1 - We need to know how to write the equation that represents the neutralization reaction that occurred and balance it correctly. Neutralization reactions follow the following scheme: acid + base -> salt + water.

In this case, the acid is nitric acid (HNO3) and base is NaOH.

The equation is:

HNO3 + NaOH -> NaNO3 + H2O

Note that the equation is already balanced. Balancing the equation correctly is important because we need to look at the stoichiometric ratio between the reactants, which in this case is 1 : 1.

Step 2 - We need to relate the obtained data. Since a total neutralization reaction took place, this means that the number of H+ ions and OH- ions present in the solution became equal. So we have:

nHNO3 = nNaOH

where n is the number os moles.

Knowing that n = M . V, where:

M is the concentration in mol/L and V is the volume in mL, we can perform the following substitution:

MHNO3 . VHNO3 = MNaOH . VNaOH

2.75 x VHNO3 = 2.94 x 486.8

VHNO3 = 520.4 mL

Note that it was not necessary to change the volume units from mL to L because they cancel out in the formula.

Answer: VHNO3 = 520.4 mL

User Incongruous
by
5.0k points