What is `DeltapH` ( final `-` initial `)` for `1//3 & 2//3` stages of neutralization of `0.1 M CH_(3)COOH` with `0.1 M NaOH`

Options

A. `+2 log 2`
B. `-2 log 3`
C. `2 log 1//4`
D. `2 log 2//3`

Explanation

To solve the problem of calculating the change in pH (ΔpH) during the neutralization of 0.1 M CH₃COOH (acetic acid) with 0.1 M NaOH at 1/3 and 2/3 stages, we can follow these steps:

### Step 1: Understand the Reaction
The neutralization reaction between acetic acid (CH₃COOH) and sodium hydroxide (NaOH) can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 2: Calculate Initial pH of Acetic Acid
To find the initial pH of the acetic acid solution, we can use the formula for the pH of a weak acid:
\[ \text{pH} = -\log[H^+] \]
For acetic acid, we need to use its dissociation constant (Ka). The dissociation of acetic acid can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
Given that the concentration of acetic acid is 0.1 M, we can use the formula:
\[ \text{Ka} = 1.8 \times 10^{-5} \]
Using the expression for Ka:
\[ \text{Ka} = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \]
Assuming that x is the concentration of H⁺ ions produced, we can set up the equation:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.1 - x} \]
For weak acids, we can assume that x is small compared to 0.1, so:
\[ 1.8 \times 10^{-5} \approx \frac{x^2}{0.1} \]
Solving for x:
\[ x^2 = 1.8 \times 10^{-6} \]
\[ x = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \]
Thus, the initial concentration of H⁺ ions is approximately \( 1.34 \times 10^{-3} \) M.
Now, calculate the initial pH:
\[ \text{pH} = -\log(1.34 \times 10^{-3}) \approx 2.87 \]

### Step 3: Calculate pH at 1/3 Stage of Neutralization
At the 1/3 stage, we have neutralized 1/3 of the acetic acid. The moles of acetic acid remaining will be:
\[ \text{Moles of CH}_3\text{COOH} = 0.1 \, \text{M} \times \frac{2}{3} \, \text{L} = \frac{0.2}{3} \, \text{mol} \]
The moles of acetate ion produced will be:
\[ \text{Moles of CH}_3\text{COO}^- = 0.1 \, \text{M} \times \frac{1}{3} \, \text{L} = \frac{0.1}{3} \, \text{mol} \]

Using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Where:
- pKa = -log(1.8 × 10⁻⁵) ≈ 4.74
- [A⁻] = concentration of acetate = \( \frac{0.1/3}{\text{total volume}} \)
- [HA] = concentration of acetic acid remaining = \( \frac{0.2/3}{\text{total volume}} \)

Assuming the total volume is 1 L after mixing:
\[ \text{pH}_{1/3} = 4.74 + \log\left(\frac{0.1/3}{0.2/3}\right) = 4.74 + \log(0.5) \approx 4.74 - 0.301 \approx 4.44 \]

### Step 4: Calculate pH at 2/3 Stage of Neutralization
At the 2/3 stage, we have neutralized 2/3 of the acetic acid:
\[ \text{Moles of CH}_3\text{COOH} = 0.1 \, \text{M} \times \frac{1}{3} \, \text{L} = \frac{0.1}{3} \, \text{mol} \]
The moles of acetate ion produced will be:
\[ \text{Moles of CH}_3\text{COO}^- = 0.1 \, \text{M} \times \frac{2}{3} \, \text{L} = \frac{0.2}{3} \, \text{mol} \]

Using the Henderson-Hasselbalch equation again:
\[ \text{pH}_{2/3} = 4.74 + \log\left(\frac{0.2/3}{0.1/3}\right) = 4.74 + \log(2) \approx 4.74 + 0.301 \approx 5.04 \]

### Step 5: Calculate ΔpH
Now we can find the change in pH:
\[ \Delta \text{pH} = \text{pH}_{2/3} - \text{pH}_{1/3} = 5.04 - 4.44 = 0.60 \]

### Final Answer
The change in pH (ΔpH) for the 1/3 and 2/3 stages of neutralization of 0.1 M CH₃COOH with 0.1 M NaOH is **0.60**.