Answer:
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
