How many mL of 0.125 M Ba(OH)2 would be required to completely neutralize 75.0 mL of 0.845 M HCl? What is the pH of the solution at the equivalence point?

253.5mL of Ba(OH)₂ are required to neutralize the HCl solution

The pH at equivalence point is = 7

Explanation:

The reaction of Ba(OH)₂ with HCl is:

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O

First, we need to find moles of HCl. With these moles and the chemical equation we can find the moles of Ba(OH)₂ and the volume required:

Moles HCl:

75.0mL = 0.075L * (0.845mol / 1L) = 0.063375moles HCl

Moles Ba(OH)₂:

0.06338moles HCl * (1mol Ba(OH)₂ / 2mol HCl) = 0.03169 moles Ba(OH)₂

Volume of the 0.125M Ba(OH)₂:

0.03169 moles Ba(OH)₂ * (1L / 0.125mol) = 0.2535L are required =

253.5mL of Ba(OH)₂ are required to neutralize the HCl solution

As the titration was of a strong acid, HCl, with a strong base, Ba(OH)₂, the pH at equivalence point is = 7

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.