A solution containing `0.2` mole of dicholoracetice acid `(K_(a)=5xx10^(-2))` and0.1 mole sodium dicholoroacetate in one litre solution has `[H^(+)]` :

Options

A. `0.05M`
B. `0.025M`
C. `0.10M`
D. `0.005M`

Explanation

To find the concentration of hydrogen ions \([H^+]\) in a solution containing 0.2 moles of dichloroacetic acid and 0.1 moles of sodium dichloroacetate, we can use the Henderson-Hasselbalch equation or set up an equilibrium expression. Here’s a step-by-step solution:

### Step 1: Identify the components
We have:
- Dichloroacetic acid \((CHCl_2COOH)\) with \(K_a = 5 \times 10^{-2}\)
- Sodium dichloroacetate \((CHCl_2COONa)\)

### Step 2: Write the dissociation equations
1. **Dichloroacetic acid dissociation**:
\[ CHCl_2COOH \rightleftharpoons CHCl_2COO^- + H^+ \]

2. **Sodium dichloroacetate dissociation** (fully dissociates):
\[ CHCl_2COONa \rightarrow CHCl_2COO^- + Na^+ \]

### Step 3: Initial concentrations
- Initial concentration of \(CHCl_2COOH\) = \(0.2 \, \text{mol/L}\)
- Initial concentration of \(CHCl_2COO^-\) from sodium dichloroacetate = \(0.1 \, \text{mol/L}\)

### Step 4: Set up the equilibrium expression
Let \(x\) be the amount of \(H^+\) produced at equilibrium. The equilibrium concentrations will be:
- \([CHCl_2COOH] = 0.2 - x\)
- \([CHCl_2COO^-] = 0.1 + x\)
- \([H^+] = x\)

### Step 5: Write the expression for \(K_a\)
Using the equilibrium concentrations, we can write the expression for \(K_a\):
\[ K_a = \frac{[CHCl_2COO^-][H^+]}{[CHCl_2COOH]} = \frac{(0.1 + x)(x)}{(0.2 - x)} \]
Given \(K_a = 5 \times 10^{-2}\), we can set up the equation:
\[ 5 \times 10^{-2} = \frac{(0.1 + x)(x)}{(0.2 - x)} \]

### Step 6: Solve for \(x\)
Multiplying both sides by \((0.2 - x)\):
\[ 5 \times 10^{-2} (0.2 - x) = (0.1 + x)(x) \]
Expanding and rearranging gives us a quadratic equation:
\[ 5 \times 10^{-2} \cdot 0.2 - 5 \times 10^{-2} x = 0.1x + x^2 \]
\[ 0.01 - 0.05x = 0.1x + x^2 \]
\[ x^2 + 0.15x - 0.01 = 0 \]

### Step 7: Apply the quadratic formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- \(a = 1\), \(b = 0.15\), \(c = -0.01\)
\[ x = \frac{-0.15 \pm \sqrt{(0.15)^2 - 4 \cdot 1 \cdot (-0.01)}}{2 \cdot 1} \]
\[ x = \frac{-0.15 \pm \sqrt{0.0225 + 0.04}}{2} \]
\[ x = \frac{-0.15 \pm \sqrt{0.0625}}{2} \]
\[ x = \frac{-0.15 \pm 0.25}{2} \]
Calculating the two possible values:
1. \(x = \frac{0.1}{2} = 0.05\) (valid)
2. \(x = \frac{-0.4}{2} = -0.2\) (not valid)

### Step 8: Conclusion
Thus, the concentration of hydrogen ions \([H^+]\) in the solution is:
\[ [H^+] = x = 0.05 \, \text{mol/L} \]