Calculate `[H^+] and [CHCl_2COO^-]` in a solution that is 0.01 M in HCl and 0.01 M in `CHCl_2COOH. "Take "(K_a= 2.55 xx 10^(-2))`.

Answer

To solve the problem of calculating the concentrations of \([H^+]\) and \([CHCl_2COO^-]\) in a solution that is 0.01 M in HCl and 0.01 M in \(CHCl_2COOH\), we can follow these steps:

### Step 1: Identify the Strong Acid and Weak Acid
- **Given**:
- HCl is a strong acid and will completely dissociate in solution.
- \(CHCl_2COOH\) is a weak acid and will partially dissociate.

### Step 2: Determine the Concentration of \(H^+\) from HCl
- Since HCl is a strong acid, it will dissociate completely:
\[ HCl \rightarrow H^+ + Cl^- \]
- Therefore, the concentration of \(H^+\) from HCl is:
\[ [H^+] = 0.01 \, M \]

### Step 3: Set Up the Dissociation Equation for \(CHCl_2COOH\)
- The dissociation of the weak acid \(CHCl_2COOH\) can be represented as:
\[ CHCl_2COOH \rightleftharpoons CHCl_2COO^- + H^+ \]
- Let \(x\) be the amount that dissociates. Therefore:
- Initial concentrations:
- \([CHCl_2COOH] = 0.01 \, M\)
- \([CHCl_2COO^-] = 0\)
- \([H^+] = 0.01 \, M\)
- At equilibrium:
- \([CHCl_2COOH] = 0.01 - x\)
- \([CHCl_2COO^-] = x\)
- \([H^+] = 0.01 + x\)

### Step 4: Write the Expression for \(K_a\)
- The expression for the acid dissociation constant \(K_a\) is given by:
\[ K_a = \frac{[CHCl_2COO^-][H^+]}{[CHCl_2COOH]} \]
- Substituting the equilibrium concentrations into the equation:
\[ K_a = \frac{x(0.01 + x)}{0.01 - x} \]
- Given \(K_a = 2.55 \times 10^{-2}\), we can set up the equation:
\[ 2.55 \times 10^{-2} = \frac{x(0.01 + x)}{0.01 - x} \]

### Step 5: Solve the Equation
- Rearranging gives:
\[ 2.55 \times 10^{-2} (0.01 - x) = x(0.01 + x) \]
- Expanding and rearranging leads to:
\[ 2.55 \times 10^{-4} - 2.55 \times 10^{-2}x = 0.01x + x^2 \]
- Rearranging all terms to one side gives:
\[ x^2 + (0.01 + 2.55 \times 10^{-2})x - 2.55 \times 10^{-4} = 0 \]
- Simplifying:
\[ x^2 + 0.036x - 2.55 \times 10^{-4} = 0 \]

### Step 6: Use the Quadratic Formula
- The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \(a = 1\), \(b = 0.036\), and \(c = -2.55 \times 10^{-4}\).
- Calculating the discriminant:
\[ b^2 - 4ac = (0.036)^2 - 4(1)(-2.55 \times 10^{-4}) \]
- Solving gives:
\[ x \approx 0.0061 \, M \]

### Step 7: Calculate \([CHCl_2COO^-]\) and \([H^+]\)
- \([CHCl_2COO^-] = x = 0.0061 \, M\)
- \([H^+] = 0.01 + x = 0.01 + 0.0061 = 0.0161 \, M\)

### Final Results
- \([H^+] \approx 0.0161 \, M\)
- \([CHCl_2COO^-] \approx 0.0061 \, M\)