The pH of a solution prepared by mixing 2.0 mL of HCI solution of pH 3.0 and 3.0 mL of NaOH of pH 10.0 is

Options

A. 2.5
B. 3.5
C. 5.5
D. 6.5

Answer

To find the pH of the solution prepared by mixing 2.0 mL of HCl solution with a pH of 3.0 and 3.0 mL of NaOH solution with a pH of 10.0, follow these steps:

### Step 1: Calculate the concentration of H⁺ ions from HCl
- The pH of the HCl solution is 3.0.
- Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
We can find [H⁺]:
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Calculate the moles of H⁺ from HCl
- Volume of HCl = 2.0 mL = 0.002 L
- Moles of H⁺ from HCl:
\[ \text{Moles of } H^+ = [H^+] \times \text{Volume} = 10^{-3} \times 0.002 = 2 \times 10^{-6} \, \text{moles} \]

### Step 3: Calculate the concentration of OH⁻ ions from NaOH
- The pH of the NaOH solution is 10.0.
- First, find the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 10 = 4 \]
- Now, calculate [OH⁻]:
\[ [OH^-] = 10^{-\text{pOH}} = 10^{-4} \, \text{M} \]

### Step 4: Calculate the moles of OH⁻ from NaOH
- Volume of NaOH = 3.0 mL = 0.003 L
- Moles of OH⁻ from NaOH:
\[ \text{Moles of } OH^- = [OH^-] \times \text{Volume} = 10^{-4} \times 0.003 = 3 \times 10^{-7} \, \text{moles} \]

### Step 5: Determine the limiting reactant
- The reaction between H⁺ and OH⁻ is:
\[ H^+ + OH^- \rightarrow H_2O \]
- We have:
- Moles of H⁺ = \(2 \times 10^{-6}\)
- Moles of OH⁻ = \(3 \times 10^{-7}\)

Since \(3 \times 10^{-7}\) moles of OH⁻ is less than \(2 \times 10^{-6}\) moles of H⁺, OH⁻ is the limiting reactant.

### Step 6: Calculate the remaining moles of H⁺ after neutralization
- Moles of H⁺ remaining after reaction:
\[ \text{Remaining } H^+ = 2 \times 10^{-6} - 3 \times 10^{-7} = 1.7 \times 10^{-6} \, \text{moles} \]

### Step 7: Calculate the total volume of the solution
- Total volume = 2.0 mL + 3.0 mL = 5.0 mL = 0.005 L

### Step 8: Calculate the concentration of remaining H⁺ ions
- Concentration of remaining H⁺:
\[ [H^+] = \frac{\text{Remaining moles}}{\text{Total volume}} = \frac{1.7 \times 10^{-6}}{0.005} = 3.4 \times 10^{-4} \, \text{M} \]

### Step 9: Calculate the pH of the final solution
- Finally, calculate the pH:
\[ \text{pH} = -\log[H^+] = -\log(3.4 \times 10^{-4}) \approx 3.46 \]

### Final Answer:
The pH of the solution is approximately **3.46**.

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