To find the concentration of hydroxide ions \([OH^-]\) in the final solution prepared by mixing \(20.0 \, \text{mL}\) of \(0.050 \, \text{M} \, \text{HCl}\) with \(30.0 \, \text{mL}\) of \(0.10 \, \text{M} \, \text{Ba(OH)}_2\), we can follow these steps:
### Step 1: Calculate the moles of HCl
To find the moles of HCl, we use the formula:
\[
\text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)}
\]
Convert the volume from mL to L:
\[
20.0 \, \text{mL} = 0.0200 \, \text{L}
\]
Now calculate the moles:
\[
\text{Moles of HCl} = 0.050 \, \text{M} \times 0.0200 \, \text{L} = 0.0010 \, \text{moles}
\]
### Step 2: Calculate the moles of Ba(OH)₂
Next, we calculate the moles of Ba(OH)₂:
\[
\text{Moles of Ba(OH)}_2 = \text{Molarity} \times \text{Volume (L)}
\]
Convert the volume from mL to L:
\[
30.0 \, \text{mL} = 0.0300 \, \text{L}
\]
Now calculate the moles:
\[
\text{Moles of Ba(OH)}_2 = 0.10 \, \text{M} \times 0.0300 \, \text{L} = 0.0030 \, \text{moles}
\]
### Step 3: Calculate the moles of OH⁻ produced
Since each Ba(OH)₂ produces 2 moles of OH⁻, we multiply the moles of Ba(OH)₂ by 2:
\[
\text{Moles of OH}^- = 2 \times 0.0030 \, \text{moles} = 0.0060 \, \text{moles}
\]
### Step 4: Determine the limiting reagent
Now we compare the moles of H⁺ from HCl and OH⁻ from Ba(OH)₂:
- Moles of H⁺ from HCl = 0.0010 moles
- Moles of OH⁻ from Ba(OH)₂ = 0.0060 moles
Since \(0.0010 < 0.0060\), H⁺ is the limiting reagent.
### Step 5: Calculate the remaining moles of OH⁻ after reaction
The moles of OH⁻ that react with H⁺ are equal to the moles of H⁺:
\[
\text{Moles of OH}^- \text{ consumed} = 0.0010 \, \text{moles}
\]
Remaining moles of OH⁻:
\[
\text{Remaining OH}^- = 0.0060 - 0.0010 = 0.0050 \, \text{moles}
\]
### Step 6: Calculate the final volume of the solution
The total volume of the mixed solution is:
\[
\text{Total Volume} = 20.0 \, \text{mL} + 30.0 \, \text{mL} = 50.0 \, \text{mL} = 0.0500 \, \text{L}
\]
### Step 7: Calculate the concentration of OH⁻
Finally, we calculate the concentration of OH⁻ in the final solution:
\[
[OH^-] = \frac{\text{Moles of OH}^-}{\text{Total Volume (L)}} = \frac{0.0050 \, \text{moles}}{0.0500 \, \text{L}} = 0.100 \, \text{M}
\]
### Final Answer
The concentration of \([OH^-]\) in the final solution is \(0.100 \, \text{M}\).
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