3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of t

Q: Why do we calculate moles instead of molarity for the reactions?

Because the volume of the solution changes when we mix different solutions. Moles are conserved during mixing, making it much easier to track the stoichiometry of the neutralization reactions without constantly recalculating concentrations.

Q: Why does NaOH react with HCl first and not CH3COOH?

HCl is a strong acid and is completely dissociated into H+ ions, whereas CH3COOH is a weak acid and is mostly unionized. The strong base NaOH will rapidly consume the readily available H+ ions from the strong acid before it forces the weak acid to dissociate further.

Q: Can we use millimoles directly in the Henderson-Hasselbalch equation?

Yes! The term $\log([\text{Salt}]/[\text{Acid}])$ is a ratio of concentrations. Since both the salt and the acid are in the same total volume, the volume terms cancel out: $(n_{\text{salt}}/V) / (n_{\text{acid}}/V) = n_{\text{salt}} / n_{\text{acid}}$. Therefore, the ratio of millimoles is exactly the same as the ratio of molarities.

Q: What if the amount of NaOH added was more than enough to neutralize both acids?

If NaOH was in excess, all the HCl and CH3COOH would be neutralized, and you would be left with a mixture of NaCl, CH3COONa, and unreacted NaOH. In that case, the pH would be determined primarily by the concentration of the strong base, excess NaOH.

Q: Why did we ignore the volume of NaOH (0.5 mL) added to the 20 mL aliquot?

The question explicitly states 'Neglect any changes in volume.' Even if it didn't, adding 0.5 mL to 20 mL is a very small change (2.5%). More importantly, as explained earlier, the Henderson-Hasselbalch equation relies on the ratio of moles, so the final total volume doesn't actually affect the pH calculation for a buffer.

Question

3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution $\frac{1}{2}$ mL of 5 M NaOH is added. The pH of the solution is …………
[Given : $pK_a$ of acetic acid = 4.75, molar mass of acetic acid = 60 g/mol, $\log 3 = 0.4771$]
Neglect any changes in volume.

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Acids

Imagine you are a chemist orchestrating a delicate dance of molecules. We start with a beaker containing a mixture of two very different characters: **acetic acid** ($	ext{CH}_3	ext{COOH}$), a weak and hesitant acid, and **hydrochloric acid** ($	ext{HCl}$), a strong and aggressive one.

First, we need to know exactly how many of these molecules we have. We are given $3	ext{ g}$ of acetic acid. Using its molar mass of $60	ext{ g/mol}$, we find we have $\frac{3}{60} = 0.05	ext{ mol}$, or **$50	ext{ mmol}$**. For the $	ext{HCl}$, we have $250	ext{ mL}$ of a $0.1	ext{ M}$ solution, which gives us $250 	imes 0.1 = $ **$25	ext{ mmol}$**.

We then dilute this entire mixture with water until the total volume reaches $500	ext{ mL}$.

### Taking a Slice: The Aliquot Principle

Now, we don't want to work with the whole $500	ext{ mL}$ batch. Instead, we carefully extract a small **$20	ext{ mL}$ aliquot**.

What happens to our moles? They scale down proportionally! The fraction of the solution we took is $\frac{20}{500} = \frac{1}{25}$. Therefore, the moles of each acid in our small beaker will also be exactly $\frac{1}{25}$th of the original amount.

For acetic acid: $50	ext{ mmol} 	imes \frac{1}{25} = $ **$2	ext{ mmol}$**.
For $	ext{HCl}$: $25	ext{ mmol} 	imes \frac{1}{25} = $ **$1	ext{ mmol}$**.

### The Neutralization Battlefield: Strong vs. Weak

Into this small beaker, we introduce a strong base: **Sodium Hydroxide** ($	ext{NaOH}$). We add $0.5	ext{ mL}$ of a $5	ext{ M}$ solution, which means we are dropping in $5 	imes 0.5 = $ **$2.5	ext{ mmol}$** of $	ext{NaOH}$.

Here is where the chemistry gets exciting. We have a strong base entering a room with both a strong acid and a weak acid. Who does it fight first? **The strong acid always wins the race.** The $	ext{NaOH}$ will preferentially and completely neutralize the $	ext{HCl}$ before it even looks at the acetic acid.

We have $1	ext{ mmol}$ of $	ext{HCl}$. It will consume exactly $1	ext{ mmol}$ of $	ext{NaOH}$ to form neutral $	ext{NaCl}$ and water.

How much $	ext{NaOH}$ survives this first skirmish? 
$2.5	ext{ mmol (initial)} - 1.0	ext{ mmol (used)} = $ **$1.5	ext{ mmol}$ of $	ext{NaOH}$ remaining.**

### The Aftermath: Birth of a Buffer

Now, the remaining $1.5	ext{ mmol}$ of $	ext{NaOH}$ turns its attention to the weak acetic acid. We started with $2	ext{ mmol}$ of acetic acid. The $	ext{NaOH}$ will react with it to form sodium acetate ($	ext{CH}_3	ext{COONa}$), which is the conjugate base of our weak acid.

Since we have $1.5	ext{ mmol}$ of $	ext{NaOH}$, it will consume exactly $1.5	ext{ mmol}$ of acetic acid.

Let's take inventory of what is left in the beaker:
- **Unreacted Acetic Acid:** $2.0 - 1.5 = $ **$0.5	ext{ mmol}$**
- **Newly Formed Sodium Acetate (Salt):** **$1.5	ext{ mmol}$**

Whenever a solution contains a significant amount of a weak acid and its conjugate base, it transforms into an **Acidic Buffer**.

### The Final Calculation: Henderson-Hasselbalch to the Rescue

To find the pH of an acidic buffer, we deploy the legendary **Henderson-Hasselbalch equation**:

$$pH = pK_a + \log \left( \frac{[	ext{Salt}]}{[	ext{Acid}]} ight)$$

Because both the salt and the acid are floating in the exact same volume of water, their volume terms cancel out perfectly. We can just plug in the ratio of their millimoles!

$$pH = 4.75 + \log \left( \frac{1.5}{0.5} ight)$$
$$pH = 4.75 + \log(3)$$

The problem kindly provides the value of $\log 3$ as $0.4771$.

$$pH = 4.75 + 0.4771 = 5.2271$$

Rounding to two decimal places, we arrive at our final, elegant answer: **$5.23$**.

Enter Numerical Value:

Visualized Solution

\text{Understanding the Setup}

\text{Initial Moles in 500 mL}

\text{Moles in 20 mL Aliquot}

\text{Moles of NaOH Added}

\text{Neutralization of Strong Acid}

\text{Neutralization of Weak Acid}

\text{Formation of Acidic Buffer}

\text{Calculating the pH}

\text{Final Answer}

The Sigma Insight: pH, Buffer and Indicator

The Setup

Taking a Slice

The Neutralization Battlefield

The Aftermath

The Final Calculation

Similar Questions

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Comprehension Passage

Enthalpy of dissociation (in $kJ mol^{- 1}$ ) of acetic acid obtained from the Expt-2 is

The $pH$ of the solution after Expt-2

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