Equal volumes of 0.2M HCI and 0.4M KOH are mixed. The concentration of ions in the resulting solution are:

Options

A. `[K^(+) = 0.4 M , [C1^(-) ] = 0.2 M , [H^(+)] = 0.2 M`
B. `[K^(+) ] = 0.2 M , [C1^(-)]=0.1M, [OH^(-)] = 0.1 M`
C. `[K^(+)]=0.1 M, [C1^(-)]=0.1 M, [OH^(-)]=0.1M`
D. `[K^(+)]=0.2M, [C1^(-)]=0.1M, [OH^(-)]=0.2M`

Answer

To find the concentration of ions in the resulting solution when equal volumes of 0.2 M HCl and 0.4 M KOH are mixed, we can follow these steps:

### Step 1: Determine the number of moles of HCl and KOH
Assuming we mix equal volumes of 1 liter of each solution:

- **For HCl:**
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{mol/L} \times 1 \, \text{L} = 0.2 \, \text{moles} \]

- **For KOH:**
\[ \text{Moles of KOH} = \text{Molarity} \times \text{Volume} = 0.4 \, \text{mol/L} \times 1 \, \text{L} = 0.4 \, \text{moles} \]

### Step 2: Write the neutralization reaction
The reaction between HCl and KOH can be written as:
\[ \text{HCl} + \text{KOH} \rightarrow \text{KCl} + \text{H}_2\text{O} \]

### Step 3: Determine the limiting reactant
From the reaction, we see that 1 mole of HCl reacts with 1 mole of KOH.

- We have 0.2 moles of HCl and 0.4 moles of KOH.
- HCl is the limiting reactant because it will be completely consumed first.

### Step 4: Calculate the remaining moles after the reaction
- **Moles of HCl remaining:**
\[ 0.2 - 0.2 = 0 \, \text{moles} \]
- **Moles of KOH remaining:**
\[ 0.4 - 0.2 = 0.2 \, \text{moles} \]

### Step 5: Calculate the moles of the products formed
- **Moles of KCl formed:**
\[ 0.2 \, \text{moles} \, \text{(from the reaction)} \]
- **Moles of water formed:**
\[ 0.2 \, \text{moles} \, \text{(from the reaction)} \]

### Step 6: Calculate the total volume of the solution
Since we mixed 1 liter of HCl and 1 liter of KOH:
\[ \text{Total volume} = 1 \, \text{L} + 1 \, \text{L} = 2 \, \text{L} \]

### Step 7: Calculate the concentration of OH⁻ ions
Since 0.2 moles of KOH remain, and 1 mole of KOH gives 1 mole of OH⁻:
- Moles of OH⁻ from KOH:
\[ 0.2 \, \text{moles} \]
- Concentration of OH⁻:
\[ \text{Concentration of OH}^- = \frac{\text{Moles of OH}^-}{\text{Total Volume}} = \frac{0.2 \, \text{moles}}{2 \, \text{L}} = 0.1 \, \text{M} \]

### Step 8: Calculate the concentration of K⁺ ions
- Moles of K⁺ from KOH:
\[ 0.2 \, \text{moles} \, \text{(remaining KOH)} \]
- Moles of K⁺ from KCl:
\[ 0.2 \, \text{moles} \, \text{(from the reaction)} \]
- Total moles of K⁺:
\[ 0.2 + 0.2 = 0.4 \, \text{moles} \]
- Concentration of K⁺:
\[ \text{Concentration of K}^+ = \frac{0.4 \, \text{moles}}{2 \, \text{L}} = 0.2 \, \text{M} \]

### Step 9: Calculate the concentration of Cl⁻ ions
- Moles of Cl⁻ from KCl:
\[ 0.2 \, \text{moles} \, \text{(from the reaction)} \]
- Concentration of Cl⁻:
\[ \text{Concentration of Cl}^- = \frac{0.2 \, \text{moles}}{2 \, \text{L}} = 0.1 \, \text{M} \]

### Final Results
- Concentration of OH⁻: 0.1 M
- Concentration of K⁺: 0.2 M
- Concentration of Cl⁻: 0.1 M