A 1 liter solution contains 0.436 M hypochlorous acid and 0.581 M potassium hypochlorite. Addition of 0.479 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)

Exceed the buffer capacity and Raise the pH by several units

Explanation:

Options are:

Raise the pH slightly

Lower the pH slightly

Raise the pH by several units

Lower the pH by several units

Not change the pH

Exceed the buffer capacity

The hypochlorous acid, HClO, is in equilibrium with Hypochlorite ion (From potassium hypochlorite, ClO⁻) producing a buffer. Using H-H equation, pH of initial buffer is:

pH = pKa + log [ClO⁻] / [HClO]

pKa for hypochlorous acid is 7.53

pH = 7.53 + log [0.581M] / [0.436M]

pH = 7.65

Barium hydroxide reacts with HClO producing more ClO⁻, thus:

Ba(OH)₂ + 2HClO →  2ClO⁻ + 2H₂O

As 0.479 moles of Barium hdroxide are added. For a complete reaction you require 0.479mol * 2 = 0.958 moles of HClO

As you have just 0.436 moles (Volume = 1L),

The addition will:

The Ba(OH)₂ that reacts is:

0.436 moles HClO * (1mole (Ba(OH)₂ / 2 mol HClO) = 0.218 moles Ba(OH)₂ and will remain:

0.479 mol - 0.218 mol = 0.261 moles Ba(OH)₂

As 1 mole of Ba(OH)₂ contains 2 moles of OH⁻, moles of OH⁻ and molarity is:

0.261 moles* 2 = 0.522 moles OH⁻ = [OH⁻]

pOH = -log [OH⁻]

pOH = 0.28

And pH = 14 - pOH:

pH = 13.72

Thus, after the addition the pH change from 7.65 to 13.62: