Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. a. 0.0 mL
b. 80.0 mL
c. 10.0 mL
d. 100.0 mL
e. 40.0 mL

a) 0.70

b) 7.00

c) 0.85

d) 12.15

e) 1.30

Explanation:

The neutralization reaction involved in the titration is:

HClO₄(aq) + KOH(aq) → KClO₄(aq) + H₂O(l)

According to the chemical equation, 1 mol of HClO₄ reacts with 1 mol of KOH (1 equivalent of acid with 1 equivalent of base). The moles are calculated from the product of the molar concentration (M) and the volume in liters.

We have the following moles of acid (HClO₄):

40.0 mL x 1 L/1000 mL = 0.04 L

0.200 mol/L x 0.04 L = 8 x 10⁻³ moles HClO₄

Since HClO₄ is a strong acid (completely dissociated into H⁺ and ClO₄⁻ ions), the moles of HClO₄ are equal to the moles of H⁺. Then, we can calculate the initial pH:

[H⁺] = 0.200 M → pH = -log [H⁺] = -log (0.200) = 0.70

Now, we calculate the pH after the addition of KOH. Since KOH is a strong base, the concentration of KOH is equal to the concentration of OH⁻ ions.

a) 0.0 mL

No KOH is added, so the pH is the initial pH: 0.70

b) 80.0 mL KOH

80.0 mL x 1 L/1000 mL = 0.08 L

0.100 mol/L x 0.08 L = 8 x 10⁻³ moles KOH = 8 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 8 x 10⁻³ moles OH⁻ = 0

The neutralization reaction is complete and there is no remaining H⁺ from the acid. The concentration of H⁺ is equal to the concentration of H⁺ of water:

[H⁺] = 1 x 10⁻⁷ M → pH = -log [H⁺] = -log (1 x 10⁻⁷) = 7.0

c) 10.0 mL KOH

10.0 mL x 1 L/1000 mL = 0.01 L

0.100 mol/L x 0.01 L = 1 x 10⁻³ moles KOH = 1 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 1 x 10⁻³ moles OH⁻ = 7 x 10⁻³ moles H⁺

The total volume is: V = 40.0 mL + 10.0 mL = 50 mL = 0.05 L

[H⁺] = 7 x 10⁻³ moles/0.05 L = 0.14  → pH = -log [H⁺] = -log (0.14) = 0.85

d) 100.0 mL KOH

100.0 mL x 1 L/1000 mL = 0.1 L

0.100 mol/L x 0.1 L = 0.01 moles KOH = 1 x 10⁻² moles OH⁻

After neutralization we have:

1 x 10⁻² moles OH⁻ - 8 x 10⁻³ moles H⁺ = 2 x 10⁻³ moles OH⁻

The total volume is: V = 40.0 mL + 100.0 mL = 140 mL = 0.14 L

[OH⁻] = 2 x 10⁻³ moles/0.14 L = 0.014  → pOH = -log [OH⁻] = -log (0.014) = 1.84

pH + pOH = 14 → pH = 14 - pOH = 14 - 1.84 = 12.15

e) 40.0 mL KOH

40.0 mL x 1 L/1000 mL = 0.04 L

0.100 mol/L x 0.04 L = 4 x 10⁻³ moles KOH = 4 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 4 x 10⁻³ moles OH⁻ = 4 x 10⁻³ moles H⁺

The total volume is: V = 40.0 mL + 40.0 mL = 80.0 mL = 0.08 L

[OH⁻] = 4 x 10⁻³ moles/0.08 L = 0.05 M  → pH = -log [H⁺] = -log (0.05) = 1.30

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