1 ml of 13.6 M HCl is diluted with water to give 1 litre of the solution. Calculate pH of the resulting solution.

Answer

To calculate the pH of the resulting solution after diluting 1 ml of 13.6 M HCl to 1 liter, we can follow these steps:

### Step 1: Use the dilution formula
The dilution formula is given by:
\[ M_1 V_1 = M_2 V_2 \]
Where:
- \( M_1 \) = initial molarity (13.6 M)
- \( V_1 \) = initial volume (1 ml)
- \( M_2 \) = final molarity (unknown)
- \( V_2 \) = final volume (1 liter = 1000 ml)

### Step 2: Convert volumes to the same unit
Since \( V_2 \) is in liters, we convert it to milliliters:
\[ V_2 = 1 \text{ L} = 1000 \text{ ml} \]

### Step 3: Substitute values into the dilution formula
Now we can substitute the known values into the formula:
\[ 13.6 \, \text{M} \times 1 \, \text{ml} = M_2 \times 1000 \, \text{ml} \]

### Step 4: Solve for \( M_2 \)
Rearranging the equation to solve for \( M_2 \):
\[ M_2 = \frac{13.6 \times 1}{1000} = 0.0136 \, \text{M} \]

### Step 5: Relate molarity to hydrogen ion concentration
Since HCl is a strong acid, it completely dissociates in solution:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Thus, the concentration of \( \text{H}^+ \) ions is equal to the molarity of the HCl solution:
\[ [\text{H}^+] = M_2 = 0.0136 \, \text{M} \]

### Step 6: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log [\text{H}^+] \]
Substituting the value:
\[ \text{pH} = -\log(0.0136) \]

### Step 7: Perform the logarithmic calculation
Using a calculator:
\[ \text{pH} = -\log(1.36 \times 10^{-2}) \]
This can be broken down as:
\[ \text{pH} = -(\log(1.36) + \log(10^{-2})) \]
\[ \text{pH} = -\log(1.36) + 2 \]

Calculating \( \log(1.36) \) gives approximately \( 0.1335 \):
\[ \text{pH} = -0.1335 + 2 \]
\[ \text{pH} \approx 1.8665 \]

### Final Answer
Thus, the pH of the resulting solution is approximately **1.87**.

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