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liubo4ka [24]
3 years ago
9

A 1 liter solution contains 0.436 M hypochlorous acid and 0.581 M potassium hypochlorite. Addition of 0.479 moles of barium hydr

oxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
Chemistry
1 answer:
jeka943 years ago
3 0

Answer:

Exceed the buffer capacity and Raise the pH by several units

Explanation:

<em>Options are:</em>

<em>Raise the pH slightly </em>

<em>Lower the pH slightly </em>

<em>Raise the pH by several units </em>

<em>Lower the pH by several units </em>

<em>Not change the pH </em>

<em>Exceed the buffer capacity</em>

<em />

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:

<h3>Exceed the buffer capacity</h3><h3 />

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:

<h3>Raise the pH by several units </h3>

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