Question

a. A 1 liter solution contains 0.509 M hypochlorous acid and 0.382 M potassium hypochlorite.

Addition of 0.191 moles of hydroiodic acid will:
(Assume that the volume does not change upon the addition of hydroiodic acid.)

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

b. A 1 liter solution contains 0.246 M hypochlorous acid and 0.328 M potassium hypochlorite.

Addition of 0.271 moles of sodium hydroxide will:
(Assume that the volume does not change upon the addition of sodium hydroxide.)

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

Answer 1

a.

Correct option: Lower the pH slightly

Explanation:

pK_a of HClO = 7.54

From Henderson-Hasselbalch equation,

pH = pK_a + log([base]/[acid])

or, pH = pK_a + log([NaClO]/[HClO])

or, pH = 7.54 + log(0.382/0.509)

or, pH = 7.54 - 0.125

or, pH = 7.41

Thus, pH of the buffer before the addition of HI = 7.41

Now, 1 L of 0.509 M of hypochlorous acid (HClO) = 1 L x 0.509 M = 0.509 mol of HClO

And, 1 L of 0.382 M of hypochlorite (NaClO) = 1 L x 0.382 M = 0.382 mol of NaClO

Hydroiodic acid is a strong acid. It reacts with NaClO to form HClO.

Thus, 0.191 mol of HI will react with 0.191 mol of NaClO to form 0.191 mol of HClO.

NaClO + HI HClO + NaI

Hence, the total moles of HClO after the addition of HI = (0.509 + 0.191) = 0.700 mol

Moles of NaClO after the addition of HI = (0.382 - 0.191) mol = 0.191 mol

The concentration of HClO after the addition of HI = 0.700 mol/1 L = 0.700 M

The concentration of NaClO after the addition of HI = 0.191 mol/1 L = 0.191 M

From Henderson-Hasselbalch equation,

pH = pK_a + log([base]/[acid])

or, pH = pK_a + log([NaClO]/[HClO])

or, pH = 7.54 + log(0.191/0.700)

= 7.54 - 0.564

= 6.97

Change in pH of the buffer solution after the addition of HI = (7.41 - 6.97) = 0.44

Therefore, 0.191 mol of HI lower the pH of the solution slightly.