Which of the following would decrease the pH of 25 cm`""^(2)` of a 0.01 M solution of HCl?

Options

A. The addition of Mg
B. The addition of 25 cm`""^(3)` of 0.02 M HCI
C. The addition of 25 cm`""^2` of 0.005 M HCI
D. None of these

Answer

To solve the question of which addition would decrease the pH of 25 cm³ of a 0.01 M solution of HCl, we will analyze each option step by step.

### Step 1: Understand the Initial Solution
We start with a 0.01 M HCl solution. HCl is a strong acid, which means it dissociates completely in water. Therefore, the concentration of hydrogen ions \([H^+]\) in the solution is equal to the concentration of HCl.

**Calculation:**
\[ [H^+] = 0.01 \, \text{M} \]

### Step 2: Calculate the Initial pH
The pH of a solution can be calculated using the formula:
\[ \text{pH} = -\log[H^+] \]

Substituting the concentration of hydrogen ions:
\[ \text{pH} = -\log(0.01) = 2 \]

### Step 3: Analyze Each Option
We will analyze the effect of each option on the pH of the solution.

#### Option 1: Addition of Magnesium
- Magnesium does not contribute to the acidity of the solution. Therefore, it will not decrease the pH.

#### Option 2: Addition of 25 cm³ of 0.20 M HCl
- Calculate the number of moles of HCl added:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.20 \, \text{M} \times 0.025 \, \text{L} = 0.005 \, \text{moles} \]
- The initial moles of HCl in 25 cm³ of 0.01 M solution:
\[ \text{Initial moles} = 0.01 \, \text{M} \times 0.025 \, \text{L} = 0.00025 \, \text{moles} \]
- Total moles after addition:
\[ \text{Total moles} = 0.005 + 0.00025 = 0.00525 \, \text{moles} \]
- Total volume after addition:
\[ \text{Total volume} = 25 \, \text{cm}^3 + 25 \, \text{cm}^3 = 50 \, \text{cm}^3 = 0.050 \, \text{L} \]
- New concentration of HCl:
\[ \text{New concentration} = \frac{0.00525 \, \text{moles}}{0.050 \, \text{L}} = 0.105 \, \text{M} \]
- New pH:
\[ \text{pH} = -\log(0.105) \approx 0.978 \]
This addition significantly decreases the pH.

#### Option 3: Addition of 25 cm³ of 0.005 M HCl
- Calculate the number of moles of HCl added:
\[ \text{Moles of HCl} = 0.005 \, \text{M} \times 0.025 \, \text{L} = 0.000125 \, \text{moles} \]
- Total moles after addition:
\[ \text{Total moles} = 0.00025 + 0.000125 = 0.000375 \, \text{moles} \]
- New concentration:
\[ \text{New concentration} = \frac{0.000375 \, \text{moles}}{0.050 \, \text{L}} = 0.0075 \, \text{M} \]
- New pH:
\[ \text{pH} = -\log(0.0075) \approx 2.12 \]
This addition results in a slight increase in pH.

### Conclusion
The addition of 25 cm³ of 0.20 M HCl will decrease the pH of the solution significantly, while the other options either do not affect the pH or increase it slightly.

### Final Answer
**The addition of 25 cm³ of 0.20 M HCl decreases the pH of the solution.**

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