After 20 " mL of " 0.1 M `Ba(OH))_(2)` is mixed with 10 " mL of " 0.2 M `HClO_(4)`, the concentration of `overset(ɵ)(O)H` ions is

Options

A. `2xx10^(-3)M`
B. `10^(-3)`M
C. 0.066 M
D. `overset(ɵ)(O)H` ions are completely neutralised.

Explanation

To find the concentration of hydroxide ions (OH⁻) after mixing 20 mL of 0.1 M Ba(OH)₂ with 10 mL of 0.2 M HClO₄, we can follow these steps:

### Step 1: Calculate the milliequivalents of Ba(OH)₂
Ba(OH)₂ dissociates to produce 2 OH⁻ ions per formula unit. Therefore, the milliequivalents of Ba(OH)₂ can be calculated as follows:

\[ \text{Milliequivalents of Ba(OH)₂} = \text{Volume (mL)} \times \text{Molarity (M)} \times \text{Number of OH⁻ ions} \]
\[ = 20 \, \text{mL} \times 0.1 \, \text{M} \times 2 = 4 \, \text{milliequivalents} \]

### Step 2: Calculate the milliequivalents of HClO₄
HClO₄ dissociates to produce 1 H⁺ ion per formula unit. Therefore, the milliequivalents of HClO₄ can be calculated as follows:

\[ \text{Milliequivalents of HClO₄} = \text{Volume (mL)} \times \text{Molarity (M)} \times \text{Number of H⁺ ions} \]
\[ = 10 \, \text{mL} \times 0.2 \, \text{M} \times 1 = 2 \, \text{milliequivalents} \]

### Step 3: Determine the remaining milliequivalents of OH⁻ ions
To find the remaining OH⁻ ions after the reaction, we subtract the milliequivalents of HClO₄ from the milliequivalents of Ba(OH)₂:

\[ \text{Remaining OH⁻ milliequivalents} = \text{Milliequivalents of Ba(OH)₂} - \text{Milliequivalents of HClO₄} \]
\[ = 4 \, \text{milliequivalents} - 2 \, \text{milliequivalents} = 2 \, \text{milliequivalents} \]

### Step 4: Calculate the total volume of the solution
The total volume of the mixed solution is the sum of the volumes of the two solutions:

\[ \text{Total Volume} = 20 \, \text{mL} + 10 \, \text{mL} = 30 \, \text{mL} \]

### Step 5: Calculate the normality of OH⁻ ions
Normality (N) is defined as the number of milliequivalents per liter of solution. To find the normality of OH⁻ ions:

\[ \text{Normality of OH⁻} = \frac{\text{Remaining OH⁻ milliequivalents}}{\text{Total Volume (mL)}} \]
\[ = \frac{2 \, \text{milliequivalents}}{30 \, \text{mL}} = 0.0667 \, \text{N} \]

### Step 6: Convert normality to molarity
Since the number of hydroxide ions (OH⁻) produced is 1, the molarity (M) is equal to the normality:

\[ \text{Molarity of OH⁻} = \text{Normality of OH⁻} = 0.0667 \, \text{M} \]

### Final Answer
The concentration of OH⁻ ions in the solution is approximately **0.0667 M**.

---