Determine the pH during the titration of 25.5 mL of 0.276 M hydroiodic acid by 0.105 M barium hydroxide at the following points: (1) Before the addition of any barium hydroxide .55 (2) After the addition of 16.7 mL of barium hydroxide (3) At the equivalence point (4) After adding 40.7 mL of barium hydroxide

1) before the addition of barium hydroxide

pH = -log[H⁺] = -log (0.276) = 0.559≈0.56

2)after the addition of barium hydroxide

pH = -log [H⁺] = -log(0.0857) = 1.067

3)at equivalent point, the solution will be neutral

pH = 7.0

4) after adding 40.7mL barium hydroxide

Explanation:

equation of reaction

2HCl(aq) + Ba(OH)₂(aq) ------->BaCl₂(aq) + 2H₂O(l)

1) Before the addition of barium hydroxide

concentration of HBr = 0.276M

[H⁺] = 0.276M

pH = -log[H⁺] = -log (0.276) = 0.559≈0.56

2) After adding 16.7mL barium hydroxide

moles of [OH⁻] = 16.7mL × 0.105 × 2

=3.507m mol = 3.507 × 10³mol

moles  of [H⁺] = 25.5mL × 0.276M

=7.038m mol = 7.038 × 10³mol

moles of  [H⁺] remaining = (7.038 - 3.421)m mol

= 3.617m mol = 3.617 × 10³mol

[H⁺]= = 0.0857

pH = -log [H⁺] = -log(0.0857) = 1.067

3) At equivalent point, the solution will be neutral

pH = 7.0

4) After adding 40.7mL barium hydroxide

moles of [OH⁻] = 40.7mL × 0.105M × 2

=8.547

moles of [OH⁻] remaining = 8.547 - 7.038

= 1.509m mol = 1.509 × 10³mol

pOH= -log[OH⁻]= 2.82

pH = 14 - 2.82 = 11.18

c. 75.7

i think that is the right