What will be the resultant pH, when 200 mL of an aqueous solution of `HCI(pH=2.0)` is mixed with 300 mL of an aqueous solution of `NaOH(pH=12.0)` ?

Explanation

To solve the problem of finding the resultant pH when 200 mL of an aqueous solution of HCl (pH = 2.0) is mixed with 300 mL of an aqueous solution of NaOH (pH = 12.0), we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions from HCl
Given that the pH of the HCl solution is 2.0, we can find the concentration of H⁺ ions using the formula:
\[ \text{[H⁺]} = 10^{-\text{pH}} \]
\[ \text{[H⁺]} = 10^{-2} = 0.01 \, \text{M} \]

### Step 2: Calculate the number of moles of H⁺ ions
To find the number of moles of H⁺ ions in the 200 mL (0.200 L) solution, we use the formula:
\[ \text{moles of H⁺} = \text{concentration} \times \text{volume} \]
\[ \text{moles of H⁺} = 0.01 \, \text{M} \times 0.200 \, \text{L} = 0.0020 \, \text{mol} \]

### Step 3: Calculate the concentration of OH⁻ ions from NaOH
Given that the pH of the NaOH solution is 12.0, we can find the pOH first:
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]
Now, we can find the concentration of OH⁻ ions:
\[ \text{[OH⁻]} = 10^{-\text{pOH}} = 10^{-2} = 0.01 \, \text{M} \]

### Step 4: Calculate the number of moles of OH⁻ ions
To find the number of moles of OH⁻ ions in the 300 mL (0.300 L) solution, we use:
\[ \text{moles of OH⁻} = \text{concentration} \times \text{volume} \]
\[ \text{moles of OH⁻} = 0.01 \, \text{M} \times 0.300 \, \text{L} = 0.0030 \, \text{mol} \]

### Step 5: Determine the limiting reagent and the remaining moles after reaction
The reaction between HCl and NaOH can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H₂O} \]

From the calculations:
- Moles of H⁺ from HCl = 0.0020 mol
- Moles of OH⁻ from NaOH = 0.0030 mol

Since HCl is the limiting reagent (less moles), it will completely react with an equivalent amount of NaOH:
- Moles of NaOH remaining = 0.0030 - 0.0020 = 0.0010 mol

### Step 6: Calculate the total volume of the mixed solution
The total volume after mixing is:
\[ \text{Total volume} = 200 \, \text{mL} + 300 \, \text{mL} = 500 \, \text{mL} = 0.500 \, \text{L} \]

### Step 7: Calculate the concentration of remaining OH⁻ ions
The concentration of remaining OH⁻ ions after the reaction is:
\[ \text{[OH⁻]} = \frac{\text{moles of OH⁻}}{\text{total volume}} = \frac{0.0010 \, \text{mol}}{0.500 \, \text{L}} = 0.0020 \, \text{M} \]

### Step 8: Calculate the pOH
Using the concentration of OH⁻ ions, we can find the pOH:
\[ \text{pOH} = -\log(\text{[OH⁻]}) = -\log(0.0020) \approx 2.699 \]

### Step 9: Calculate the resultant pH
Finally, we can find the resultant pH using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pH} = 14 - \text{pOH} = 14 - 2.699 \approx 11.301 \]

Thus, the resultant pH is approximately **11**.

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