To find the final pH of the solution when 4 ml of HCl solution (pH = 2) is mixed with 6 ml of NaOH solution (pH = 12), 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, we can calculate the concentration of H⁺ ions using the formula:
\[ \text{[H⁺]} = 10^{-\text{pH}} \]
\[ \text{[H⁺]} = 10^{-2} = 0.01 \, \text{mol/L} \]
### Step 2: Calculate the number of moles of H⁺ in the HCl solution
To find the number of moles of H⁺ ions in 4 ml (0.004 L) of the HCl solution:
\[ \text{moles of H⁺} = \text{concentration} \times \text{volume} \]
\[ \text{moles of H⁺} = 0.01 \, \text{mol/L} \times 0.004 \, \text{L} = 4 \times 10^{-5} \, \text{mol} \]
### Step 3: Calculate the concentration of OH⁻ ions from NaOH
Given that the pH of the NaOH solution is 12, we can find the concentration of OH⁻ ions:
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]
\[ \text{[OH⁻]} = 10^{-\text{pOH}} = 10^{-2} = 0.01 \, \text{mol/L} \]
### Step 4: Calculate the number of moles of OH⁻ in the NaOH solution
To find the number of moles of OH⁻ ions in 6 ml (0.006 L) of the NaOH solution:
\[ \text{moles of OH⁻} = \text{concentration} \times \text{volume} \]
\[ \text{moles of OH⁻} = 0.01 \, \text{mol/L} \times 0.006 \, \text{L} = 6 \times 10^{-5} \, \text{mol} \]
### Step 5: Determine the limiting reactant and calculate the excess
The reaction between HCl and NaOH is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H₂O} \]
From the calculations:
- Moles of H⁺ = \( 4 \times 10^{-5} \)
- Moles of OH⁻ = \( 6 \times 10^{-5} \)
Since 1 mole of H⁺ reacts with 1 mole of OH⁻, we can find the excess:
\[ \text{Excess OH⁻} = \text{moles of OH⁻} - \text{moles of H⁺} \]
\[ \text{Excess OH⁻} = 6 \times 10^{-5} - 4 \times 10^{-5} = 2 \times 10^{-5} \, \text{mol} \]
### Step 6: Calculate the concentration of excess OH⁻ in the final solution
The total volume of the mixed solution is:
\[ \text{Total Volume} = 4 \, \text{ml} + 6 \, \text{ml} = 10 \, \text{ml} = 0.01 \, \text{L} \]
Now, we can find the concentration of the excess OH⁻ ions:
\[ \text{[OH⁻]} = \frac{\text{Excess moles of OH⁻}}{\text{Total Volume}} \]
\[ \text{[OH⁻]} = \frac{2 \times 10^{-5}}{0.01} = 0.002 \, \text{mol/L} \]
### Step 7: Calculate pOH and then pH
Now we can calculate pOH:
\[ \text{pOH} = -\log(\text{[OH⁻]}) = -\log(0.002) \]
Using the logarithmic property:
\[ \text{pOH} = -\log(2 \times 10^{-3}) = -\log(2) - \log(10^{-3}) = -0.3 + 3 = 2.7 \]
Finally, we can find pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 2.7 = 11.3 \]
### Final Answer
The final pH of the solution is **11.3**.
