What is the `[OH]^(-)` in the final solution prepared by mixing 20.0mL of 0.050M HCl with 30.0 mL of 0.10 M Ba `(OH)_(2)`?

Options

A. 0.10 M
B. 0.40 M
C. 0.0050 M
D. 0.12 M

Explanation

To find the concentration of hydroxide ions \([OH]^-\) in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)₂, we can follow these steps:

### Step 1: Calculate the moles of HCl
First, we need to calculate the number of moles of HCl in the solution.

\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} \]

Given:
- Concentration of HCl = 0.050 M
- Volume of HCl = 20.0 mL = 0.020 L (convert mL to L)

\[ \text{Moles of HCl} = 0.050 \, \text{mol/L} \times 0.020 \, \text{L} = 0.001 \, \text{mol} \]

### Step 2: Calculate the moles of Ba(OH)₂
Next, we calculate the number of moles of Ba(OH)₂ in the solution.

\[ \text{Moles of Ba(OH)}_2 = \text{Concentration} \times \text{Volume} \]

Given:
- Concentration of Ba(OH)₂ = 0.10 M
- Volume of Ba(OH)₂ = 30.0 mL = 0.030 L

\[ \text{Moles of Ba(OH)}_2 = 0.10 \, \text{mol/L} \times 0.030 \, \text{L} = 0.003 \, \text{mol} \]

### Step 3: Calculate the moles of hydroxide ions from Ba(OH)₂
Barium hydroxide dissociates in water as follows:

\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]

This means that each mole of Ba(OH)₂ produces 2 moles of OH⁻ ions.

\[ \text{Moles of OH}^- = 2 \times \text{Moles of Ba(OH)}_2 = 2 \times 0.003 \, \text{mol} = 0.006 \, \text{mol} \]

### Step 4: Determine the limiting reactant
Now we need to determine how many moles of H⁺ ions will react with the OH⁻ ions.

The reaction between H⁺ and OH⁻ can be represented as:

\[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \]

From the previous calculations:
- Moles of H⁺ = 0.001 mol (from HCl)
- Moles of OH⁻ = 0.006 mol (from Ba(OH)₂)

Since H⁺ is the limiting reactant, it will react with an equal amount of OH⁻.

### Step 5: Calculate the remaining moles of OH⁻
After the reaction, the remaining moles of OH⁻ will be:

\[ \text{Remaining OH}^- = \text{Initial OH}^- - \text{H}^+ = 0.006 \, \text{mol} - 0.001 \, \text{mol} = 0.005 \, \text{mol} \]

### Step 6: Calculate the total volume of the solution
The total volume of the mixed solution is:

\[ \text{Total Volume} = 20.0 \, \text{mL} + 30.0 \, \text{mL} = 50.0 \, \text{mL} = 0.050 \, \text{L} \]

### Step 7: Calculate the concentration of OH⁻ ions
Finally, we can calculate the concentration of the remaining OH⁻ ions:

\[ \text{Concentration of OH}^- = \frac{\text{Moles of OH}^-}{\text{Total Volume}} = \frac{0.005 \, \text{mol}}{0.050 \, \text{L}} = 0.10 \, \text{M} \]

### Final Answer
The concentration of \([OH]^-\) in the final solution is **0.10 M**.

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