To calculate the pH of a solution made by mixing 50 mL of 0.01 M Ba(OH)₂ with 50 mL of water, we will follow these steps:
### Step 1: Calculate the moles of Ba(OH)₂
First, we need to find the number of moles of Ba(OH)₂ in the solution.
\[
\text{Molarity (M)} = \frac{\text{moles}}{\text{volume (L)}}
\]
Given:
- Volume of Ba(OH)₂ = 50 mL = 0.050 L
- Concentration of Ba(OH)₂ = 0.01 M
Now, calculate the moles:
\[
\text{Moles of Ba(OH)₂} = \text{Molarity} \times \text{Volume} = 0.01 \, \text{mol/L} \times 0.050 \, \text{L} = 0.0005 \, \text{mol}
\]
### Step 2: Determine the concentration of OH⁻ ions
Ba(OH)₂ dissociates completely in water as follows:
\[
\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^-
\]
From the dissociation, we see that 1 mole of Ba(OH)₂ produces 2 moles of OH⁻ ions. Therefore, the number of moles of OH⁻ produced is:
\[
\text{Moles of OH}^- = 2 \times \text{Moles of Ba(OH)₂} = 2 \times 0.0005 \, \text{mol} = 0.001 \, \text{mol}
\]
### Step 3: Calculate the total volume of the solution
The total volume after mixing 50 mL of Ba(OH)₂ with 50 mL of water is:
\[
\text{Total Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L}
\]
### Step 4: Calculate the concentration of OH⁻ ions in the final solution
Now, we can find the concentration of OH⁻ ions:
\[
\text{Concentration of OH}^- = \frac{\text{Moles of OH}^-}{\text{Total Volume}} = \frac{0.001 \, \text{mol}}{0.1 \, \text{L}} = 0.01 \, \text{M}
\]
### Step 5: Calculate the pOH of the solution
To find the pOH, we use the formula:
\[
\text{pOH} = -\log[\text{OH}^-]
\]
Substituting the concentration of OH⁻:
\[
\text{pOH} = -\log(0.01) = -\log(10^{-2}) = 2
\]
### Step 6: Calculate the pH of the solution
Using the relationship between pH and pOH:
\[
\text{pH} + \text{pOH} = 14
\]
We can find the pH:
\[
\text{pH} = 14 - \text{pOH} = 14 - 2 = 12
\]
### Final Answer
The pH of the solution is **12**.
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