What is the `[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)`?

Options

A. (a)`0.12 M`
B. (b)`0.10 M`
C. (c )`0.40 M`
D. (d)`0.0050 M`

Explanation

To find the concentration of hydroxide ions \([OH^-]\) in the final solution after 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
First, we need to calculate the number of moles of HCl in the solution.

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

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

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

### Step 3: Determine the moles of OH⁻ from Ba(OH)₂
Since Ba(OH)₂ dissociates into one Ba²⁺ ion and two OH⁻ ions, the moles of OH⁻ produced will be twice the moles of Ba(OH)₂.

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

### Step 4: Calculate the moles of OH⁻ remaining after neutralization
Now we need to find out how many moles of OH⁻ remain after reacting with HCl. The reaction between HCl and OH⁻ is a 1:1 reaction.

\[ \text{Moles of OH}^- \text{ remaining} = \text{Moles of OH}^- - \text{Moles of HCl} = 0.006 \, \text{mol} - 0.001 \, \text{mol} = 0.005 \, \text{mol} \]

### Step 5: Calculate the total volume of the final solution
The total volume of the final solution is the sum of the volumes of HCl and Ba(OH)₂.

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

### Step 6: Calculate the concentration of OH⁻ in the final solution
Finally, we can find the concentration of OH⁻ in the final solution.

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

Thus, the concentration of hydroxide ions \([OH^-]\) in the final solution is \(0.10 \, \text{M}\).

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