How much water is to be added to prepare 0.12 M solutions from 100 mL of 0.3 M solution?

Answer

To solve the problem of how much water needs to be added to prepare a 0.12 M solution from 100 mL of a 0.3 M solution, we can use the dilution equation:

\[ M_1 V_1 = M_2 V_2 \]

Where:
- \( M_1 \) = initial molarity (0.3 M)
- \( V_1 \) = initial volume (100 mL)
- \( M_2 \) = final molarity (0.12 M)
- \( V_2 \) = final volume (which is the initial volume plus the volume of water added)

### Step-by-step solution:

1. **Identify the known values:**
- \( M_1 = 0.3 \, \text{M} \)
- \( V_1 = 100 \, \text{mL} \)
- \( M_2 = 0.12 \, \text{M} \)

2. **Set up the equation:**
- We need to express \( V_2 \) in terms of the volume of water added. Let \( V \) be the volume of water added in mL.
- Therefore, the final volume \( V_2 = 100 \, \text{mL} + V \).

3. **Substitute the values into the dilution equation:**
\[ M_1 V_1 = M_2 V_2 \]
\[ 0.3 \times 100 = 0.12 \times (100 + V) \]

4. **Calculate the left side:**
\[ 30 = 0.12 \times (100 + V) \]

5. **Distribute \( 0.12 \) on the right side:**
\[ 30 = 12 + 0.12V \]

6. **Rearrange the equation to isolate \( V \):**
\[ 30 - 12 = 0.12V \]
\[ 18 = 0.12V \]

7. **Solve for \( V \):**
\[ V = \frac{18}{0.12} = 150 \, \text{mL} \]

8. **Conclusion:**
- Therefore, to prepare a 0.12 M solution from 100 mL of a 0.3 M solution, you need to add **150 mL of water**.