To solve this titration problem step by step, let's begin with the relevant information.
Given Information:
- Volume of acetic acid (HA): 25.0 mL
- Concentration of acetic acid (HA): 0.10 M
- Volume of NaOH solution used: varies (0 mL, then 15.0 mL, then 25.0 mL)
- Concentration of NaOH: 0.10 M
A. pH of the acid solution before NaOH is added
Step 1: Calculate the pKa of acetic acid: The dissociation reaction of acetic acid (HA) can be represented as follows:
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]
Where:
- HA = acetic acid (CH₃COOH)
- A⁻ = acetate ion (CH₃COO⁻)
For acetic acid, \( K_a \approx 1.8 \times 10^{-5} \).
Using the relationship:
\[ pK_a = -\log(K_a) \implies pK_a = -\log(1.8 \times 10^{-5}) \approx 4.76 \]
Step 2: Calculate the pH using the formula: Using the formula for weak acids, we can approximate the pH using:
\[ pH = \frac{1}{2}(pK_a - \log[C_0]) \]
Where \( C_0 = 0.10, M \):
\[ pH \approx \frac{1}{2}(4.76 - \log(0.10)) \]
Using \( \log(0.10) = -1 \):
\[ pH \approx \frac{1}{2}(4.76 + 1) = \frac{5.76}{2} \approx 2.88 \]
B. pH after adding 15.0 mL of NaOH
Step 1: Balanced equations The balanced actual equation:
\[ \text{HA}(aq) + \text{NaOH}(aq) \rightarrow \text{A}^-(aq) + \text{H}_2\text{O}(l) \]
Step 2: Initial moles and concentrations
-
Moles of acetic acid (initial): \[ \text{Moles} = C \times V = 0.10, M \times 0.025, L = 0.0025, moles \]
-
Moles of NaOH added: \[ \text{Moles} = C \times V = 0.10, M \times 0.015, L = 0.0015, moles \]
Step 3: Use ICE table
| | HA | A⁻ | H⁺ | |------------------|--------------|--------------|---------------| | Initial (mol) | 0.0025 | 0 | 0 | | Change (mol) | -0.0015 | +0.0015 | +0.0015 | | Equilibrium (mol)| 0.0010 | 0.0015 | 0.0015 |
Equilibrium concentrations (in total 40 mL):
- \[ \text{[HA]} = \frac{0.0010}{0.040} = 0.025, M \]
- \[ \text{[A⁻]} = \frac{0.0015}{0.040} = 0.0375, M \]
- \[ \text{[H⁺]} = \frac{0.0015}{0.040} = 0.0375, M \]
Step 4: Calculate pH
Using the concentration of H⁺:
\[ \text{pH} = -\log[H^+] = -\log(0.0375) \approx 1.43 \]
C. Limiting reagent and amount
The limiting reagent is acetic acid (HA). The initial amount was 0.0025 moles, and after 15 mL of NaOH added, 0.0015 moles were consumed, leaving:
\[ 0.0025 - 0.0015 = 0.0010, moles \text{ of HA remaining} \]
Thus, the limiting reagent is HA by 0.0015 moles.
D. pH after adding 25.0 mL NaOH
Step 1: Moles after 25 mL added
- Moles of NaOH: \[ \text{Moles} = 0.10, M \times 0.025, L = 0.0025, moles \]
Step 2: Reaction Stoichiometry
- Initial moles of acetic acid: 0.0025 moles
- Moles of NaOH added: 0.0025 moles
Balanced equation: \[ \text{HA} + \text{NaOH} \rightarrow \text{A}^- + \text{H}_2\text{O} \]
Change in moles:
| | HA | A⁻ | H⁺ | |------------------|----------------|----------------|----------------| | Initial (mol) | 0.0025 | 0 | 0 | | Change (mol) | -0.0025 | +0.0025 | +0 | | Equilibrium (mol)| 0 | 0.0025 | 0 |
Now, as all the acetic acid has reacted:
Step 3: Calculate concentrations in total volume of 50 mL:
- \[ \text{[A⁻]} = \frac{0.0025}{0.050} = 0.050, M \]
Step 4: Calculate pH using the \( K_b \) of the acetate ion:
\[ K_w = 1.0 \times 10^{-14} \Rightarrow K_b = \frac{K_w}{K_a} \]
\[ K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]
Using the formula for weak bases:
\[ K_b = \frac{[OH^-]^2}{[A^-]} \]
Let \( x = [OH^-] \):
\[ 5.56 \times 10^{-10} = \frac{x^2}{0.050} \] \[ x^2 = 5.56 \times 10^{-10} \times 0.050 \implies x^2 \approx 2.78 \times 10^{-11} \implies x \approx 5.27 \times 10^{-6} \]
Now, calculating pOH:
\[ pOH = -\log(5.27 \times 10^{-6}) \approx 5.28 \]
And converting to pH:
\[ pH + pOH = 14 \Rightarrow pH = 14 - 5.28 \approx 8.72 \]
E. Titration Curve Sketch
- Beginning (pH ~ 2.88): Start with acetic acid, a weak acid.
- At 15.0 mL NaOH: pH rises, partially neutralized (pH ~ 4.76).
- At 25.0 mL NaOH: Complete neutralization reached near the equivalence point resulting in a pH of ~8.72.
- After The Equivalence Point (beyond 25.0 mL): The pH would continue to rise more rapidly reflecting the addition of excess base.
This progression will be a curve, starting low and gradually climbing upward, with an inflection point around the equivalence point (approximately at 25 mL added), where the change in pH becomes very rapid.
[Insert your own sketch following these descriptions, as I'm unable to create visual diagrams.]