When a drop of a concentrated HCl solution is added to one litre of pure water at `25^(@)C`, the pH drops from 7 to about 4. When the second drop of the same HCI solution is added, the pH further drops to about

Options

A. 3.7
B. 2
C. 1
D. 0

Answer

To solve the problem step by step, we will analyze the effect of adding concentrated HCl to pure water and how it affects the pH.

### Step 1: Understand the initial condition
The initial pH of pure water at 25°C is 7. This corresponds to a hydrogen ion concentration \([H^+]\) of:
\[ [H^+] = 10^{-7} \, \text{M} \]

**Hint:** Recall that pH is defined as the negative logarithm of the hydrogen ion concentration.

### Step 2: Determine the effect of the first drop of HCl
When the first drop of concentrated HCl is added, the pH drops to 4. This means the new hydrogen ion concentration is:
\[ [H^+] = 10^{-4} \, \text{M} \]

**Hint:** Remember that a lower pH indicates a higher concentration of hydrogen ions.

### Step 3: Calculate the change in hydrogen ion concentration
The concentration of hydrogen ions after adding the first drop of HCl is \(10^{-4} \, \text{M}\). The contribution from the pure water is negligible compared to this concentration.

**Hint:** Consider that the contribution from the water's own ionization is very small compared to strong acids like HCl.

### Step 4: Adding the second drop of HCl
When the second drop of the same concentrated HCl solution is added, it will also contribute an additional \(10^{-4} \, \text{M}\) of hydrogen ions. Therefore, the total concentration of hydrogen ions after adding the second drop becomes:
\[ [H^+] = 10^{-4} + 10^{-4} = 2 \times 10^{-4} \, \text{M} \]

**Hint:** Think about how the concentrations add up when you add more of the same strong acid.

### Step 5: Calculate the new pH
To find the new pH after the second drop, we use the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the total concentration:
\[ \text{pH} = -\log(2 \times 10^{-4}) = -\log(2) - \log(10^{-4}) = -\log(2) + 4 \]
Using \(\log(2) \approx 0.3\):
\[ \text{pH} \approx 4 - 0.3 = 3.7 \]

**Hint:** Remember that the logarithm of a product is the sum of the logarithms.

### Final Answer
The pH after adding the second drop of concentrated HCl is approximately **3.7**.