What will be the resultant pH when 200 mL of an aqueous solution of HCI (pH = 2) is mixed with 300 mL of an aqueosu solution of NaOH (pH = 12)?

Options

A. 10
B. 11
C. 12
D. 14

Answer

To find the resultant pH when 200 mL of an aqueous solution of HCl (pH = 2) is mixed with 300 mL of an aqueous solution of NaOH (pH = 12), we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions from HCl

Given:
- pH of HCl = 2

Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]

We can find the concentration of H⁺ ions:
\[ [H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M} \]

### Step 2: Calculate the moles of H⁺ from HCl

Volume of HCl solution = 200 mL = 0.200 L

Moles of H⁺ from HCl:
\[ \text{Moles of H}^+ = [H^+] \times \text{Volume} = 10^{-2} \, \text{M} \times 0.200 \, \text{L} = 0.002 \, \text{moles} \]

### Step 3: Calculate the concentration of OH⁻ ions from NaOH

Given:
- pH of NaOH = 12

Using the relationship between pH and pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]

Now, we can find the concentration of OH⁻ ions:
\[ [OH^-] = 10^{-\text{pOH}} = 10^{-2} \, \text{M} \]

### Step 4: Calculate the moles of OH⁻ from NaOH

Volume of NaOH solution = 300 mL = 0.300 L

Moles of OH⁻ from NaOH:
\[ \text{Moles of OH}^- = [OH^-] \times \text{Volume} = 10^{-2} \, \text{M} \times 0.300 \, \text{L} = 0.003 \, \text{moles} \]

### Step 5: Determine the limiting reactant

The neutralization reaction between H⁺ and OH⁻ is:
\[ H^+ + OH^- \rightarrow H_2O \]

From the calculations:
- Moles of H⁺ = 0.002 moles
- Moles of OH⁻ = 0.003 moles

Since H⁺ is the limiting reactant, it will react completely with an equivalent amount of OH⁻.

### Step 6: Calculate the remaining moles of OH⁻ after reaction

Moles of OH⁻ remaining after neutralization:
\[ \text{Remaining OH}^- = \text{Initial OH}^- - \text{H}^+ = 0.003 - 0.002 = 0.001 \, \text{moles} \]

### Step 7: Calculate the total volume of the mixture

Total volume = Volume of HCl + Volume of NaOH = 200 mL + 300 mL = 500 mL = 0.500 L

### Step 8: Calculate the concentration of remaining OH⁻ ions

\[ [OH^-] = \frac{\text{Remaining moles of OH}^-}{\text{Total volume}} = \frac{0.001 \, \text{moles}}{0.500 \, \text{L}} = 0.002 \, \text{M} \]

### Step 9: Calculate the pOH and then the pH

Calculate pOH:
\[ \text{pOH} = -\log[OH^-] = -\log(0.002) \approx 2.7 \]

Now, calculate pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 2.7 = 11.3 \]

### Final Answer:
The resultant pH of the mixture is approximately **11.3**.

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