To find the pH of the solution formed by mixing `40 cm³` of `0.1 M HCl` with `10 cm³` of `0.45 M NaOH`, we can follow these steps:
### Step 1: Calculate the number of moles of HCl
- The number of moles of HCl can be calculated using the formula:
\[
\text{Number of moles} = \text{Molarity} \times \text{Volume (in L)}
\]
- Convert the volume from cm³ to L:
\[
40 \, \text{cm}^3 = 40 \, \text{mL} = \frac{40}{1000} \, \text{L} = 0.04 \, \text{L}
\]
- Now, calculate the moles of HCl:
\[
\text{Moles of HCl} = 0.1 \, \text{mol/L} \times 0.04 \, \text{L} = 0.004 \, \text{mol} = 4 \, \text{mmol}
\]
### Step 2: Calculate the number of moles of NaOH
- Similarly, calculate the number of moles of NaOH:
\[
10 \, \text{cm}^3 = 10 \, \text{mL} = \frac{10}{1000} \, \text{L} = 0.01 \, \text{L}
\]
- Now, calculate the moles of NaOH:
\[
\text{Moles of NaOH} = 0.45 \, \text{mol/L} \times 0.01 \, \text{L} = 0.0045 \, \text{mol} = 4.5 \, \text{mmol}
\]
### Step 3: Determine the limiting reactant
- HCl and NaOH react in a 1:1 ratio. We have:
- Moles of HCl = 4 mmol
- Moles of NaOH = 4.5 mmol
- Since HCl is the limiting reactant, it will be completely consumed, and we will have:
\[
\text{Remaining moles of NaOH} = 4.5 \, \text{mmol} - 4 \, \text{mmol} = 0.5 \, \text{mmol}
\]
### Step 4: Calculate the total volume of the solution
- The total volume of the solution after mixing is:
\[
\text{Total volume} = 40 \, \text{cm}^3 + 10 \, \text{cm}^3 = 50 \, \text{cm}^3 = 0.05 \, \text{L}
\]
### Step 5: Calculate the concentration of OH⁻ ions
- The concentration of NaOH (which gives OH⁻ ions) in the final solution is:
\[
\text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Total Volume}} = \frac{0.0005 \, \text{mol}}{0.05 \, \text{L}} = 0.01 \, \text{M}
\]
### Step 6: Calculate pOH and then pH
- Calculate pOH:
\[
\text{pOH} = -\log[\text{OH}^-] = -\log(0.01) = 2
\]
- Now use the relation \( \text{pH} + \text{pOH} = 14 \):
\[
\text{pH} = 14 - \text{pOH} = 14 - 2 = 12
\]
### Final Answer
The pH of the solution is **12**.
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