Which of the following statement `is//are` correct?
`20.0 mL` of `6.0 M HCl` is mixed with `50.0 mL` of `2.0 M Ba(OH)_(2)`, and `30 mL` of water is added.

Options

A. The concentration of `overset(ɵ)(OH)` remaining in solution is `0.8 M`.
B. The concentration of `Cl^(ɵ)` remaining in solution is `1.2 M`
C. The concentration of `Ba^(2+)` remaining in solution is `1.0 M`
D. 80 mmols of `overset(ɵ)(OH)` is in excess.

Explanation

To solve the problem, we will follow a systematic approach to determine the concentrations of hydroxide ions, chloride ions, and barium ions in the solution after mixing the given volumes of HCl and Ba(OH)₂, and then adding water.

### Step-by-Step Solution:

**Step 1: Calculate the moles of HCl.**
- Given: Volume of HCl = 20.0 mL = 0.020 L
- Molarity of HCl = 6.0 M
- Moles of HCl = Molarity × Volume = 6.0 mol/L × 0.020 L = 0.12 moles = 120 millimoles

**Step 2: Calculate the moles of Ba(OH)₂.**
- Given: Volume of Ba(OH)₂ = 50.0 mL = 0.050 L
- Molarity of Ba(OH)₂ = 2.0 M
- Moles of Ba(OH)₂ = Molarity × Volume = 2.0 mol/L × 0.050 L = 0.10 moles = 100 millimoles

**Step 3: Write the balanced chemical equation for the reaction.**
- The reaction between HCl and Ba(OH)₂ is:
\[ 2 \text{HCl} + \text{Ba(OH)}_2 \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O} \]

**Step 4: Determine the limiting reagent.**
- From the balanced equation, 1 mole of Ba(OH)₂ reacts with 2 moles of HCl.
- 100 millimoles of Ba(OH)₂ would require 200 millimoles of HCl.
- We only have 120 millimoles of HCl, so HCl is the limiting reagent.

**Step 5: Calculate the amount of Ba(OH)₂ that reacts.**
- Since HCl is limiting, it will react completely:
- Moles of HCl used = 120 millimoles
- Moles of Ba(OH)₂ reacted = 120 millimoles HCl × (1 mole Ba(OH)₂ / 2 moles HCl) = 60 millimoles

**Step 6: Calculate the remaining moles of Ba(OH)₂.**
- Initial moles of Ba(OH)₂ = 100 millimoles
- Moles of Ba(OH)₂ remaining = 100 millimoles - 60 millimoles = 40 millimoles

**Step 7: Calculate the moles of products formed.**
- Moles of BaCl₂ formed = Moles of HCl used / 2 = 120 millimoles / 2 = 60 millimoles

**Step 8: Calculate the total volume of the solution.**
- Total volume = Volume of HCl + Volume of Ba(OH)₂ + Volume of water
- Total volume = 20 mL + 50 mL + 30 mL = 100 mL = 0.1 L

**Step 9: Calculate the concentrations of the ions in the solution.**

1. **Concentration of hydroxide ions (OH⁻):**
- Moles of OH⁻ from remaining Ba(OH)₂ = 40 millimoles × 2 = 80 millimoles
- Concentration of OH⁻ = Moles / Volume = 80 millimoles / 100 mL = 0.8 M

2. **Concentration of chloride ions (Cl⁻):**
- Moles of Cl⁻ from BaCl₂ = 60 millimoles × 2 = 120 millimoles
- Concentration of Cl⁻ = Moles / Volume = 120 millimoles / 100 mL = 1.2 M

3. **Concentration of barium ions (Ba²⁺):**
- Moles of Ba²⁺ from remaining Ba(OH)₂ = 40 millimoles
- Moles of Ba²⁺ from BaCl₂ = 60 millimoles
- Total moles of Ba²⁺ = 40 + 60 = 100 millimoles
- Concentration of Ba²⁺ = Moles / Volume = 100 millimoles / 100 mL = 1.0 M

### Final Results:
- Concentration of OH⁻ = 0.8 M
- Concentration of Cl⁻ = 1.2 M
- Concentration of Ba²⁺ = 1.0 M