100 mL of a solution with pH = 6 is diluted to 1000 mL by adding water. pH will

Options

A. decrease by 0.7 unit
B. increase by 1 unit
C. increase by 9 units
D. increase by 0.7 unit

Answer

To solve the question step by step, we need to analyze the effect of dilution on the pH of a solution.

### Step 1: Understand the Initial Conditions
We have a solution with a volume of 100 mL and a pH of 6.

### Step 2: Calculate the Initial Concentration of H⁺ Ions
The pH is related to the concentration of hydrogen ions [H⁺] by the formula:
\[ \text{pH} = -\log [\text{H}^+] \]
Given that pH = 6, we can find [H⁺]:
\[ [\text{H}^+] = 10^{-6} \, \text{M} \]

### Step 3: Calculate the Amount of H⁺ Ions in the Initial Solution
To find the total moles of H⁺ in the initial 100 mL solution:
\[ \text{Moles of H}^+ = [\text{H}^+] \times \text{Volume} = 10^{-6} \, \text{M} \times 0.1 \, \text{L} = 10^{-7} \, \text{moles} \]

### Step 4: Dilute the Solution
The solution is diluted to a total volume of 1000 mL (1 L).

### Step 5: Calculate the New Concentration of H⁺ Ions After Dilution
Using the dilution formula, we can find the new concentration of H⁺ ions:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 = 10^{-6} \, \text{M} \) (initial concentration)
- \( V_1 = 0.1 \, \text{L} \) (initial volume)
- \( V_2 = 1 \, \text{L} \) (final volume)

Now we can solve for \( C_2 \):
\[ 10^{-6} \, \text{M} \times 0.1 \, \text{L} = C_2 \times 1 \, \text{L} \]
\[ C_2 = 10^{-6} \times 0.1 = 10^{-7} \, \text{M} \]

### Step 6: Calculate the New pH
Now we can calculate the new pH:
\[ \text{pH} = -\log [\text{H}^+] \]
Since the concentration of H⁺ after dilution is \( 10^{-7} \, \text{M} \):
\[ \text{pH} = -\log(10^{-7}) = 7 \]

### Step 7: Determine the Change in pH
The initial pH was 6, and the new pH after dilution is 7. Therefore, the change in pH is:
\[ \text{Change in pH} = 7 - 6 = 1 \]

### Conclusion
The pH of the solution increases by 1 unit after dilution.

### Final Answer
The correct option is that the pH will increase by 1 unit.

---