0.5 moles of HCI and 0.5 moles of `CH_(3)COONa` are disolved in water and the solution is made upto 500 ml. The `H ^+`of the resulting solution will be: `(K_(a) (CH_(3)COOH) = 1.6 xx 10^(-5))`

Options

A. `1.6 xx 10^(-5)`
B. `1.6 xx 10^(-4)`
C. `4 xx 10^(-3)`
D. `4 xx 10^(-2)`

Answer

To find the concentration of \( H^+ \) ions in a solution containing 0.5 moles of HCl and 0.5 moles of sodium acetate (CH₃COONa) in a total volume of 500 mL, we can follow these steps:

### Step 1: Calculate the initial concentrations of HCl and CH₃COONa
Given that we have 0.5 moles of each solute and the total volume of the solution is 500 mL (or 0.5 L), we can calculate the concentrations as follows:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in L}} \]

For both HCl and CH₃COONa:

\[ \text{Concentration of HCl} = \frac{0.5 \text{ moles}}{0.5 \text{ L}} = 1 \text{ M} \]

\[ \text{Concentration of CH₃COONa} = \frac{0.5 \text{ moles}}{0.5 \text{ L}} = 1 \text{ M} \]

### Step 2: Understand the reaction and equilibrium
When HCl (a strong acid) is added to sodium acetate (a salt of a weak acid), the following equilibrium reaction occurs:

\[ \text{HCl} + \text{CH}_3\text{COO}^- \rightleftharpoons \text{CH}_3\text{COOH} + \text{Cl}^- \]

Initially, we have:
- HCl: 1 M
- CH₃COO⁻: 1 M
- CH₃COOH: 0 M

At equilibrium, let \( x \) be the amount of HCl that reacts. Therefore, the concentrations at equilibrium will be:
- HCl: \( 1 - x \)
- CH₃COO⁻: \( 1 - x \)
- CH₃COOH: \( x \)

### Step 3: Apply the equilibrium constant expression
The equilibrium constant \( K_a \) for acetic acid (CH₃COOH) is given as \( 1.6 \times 10^{-5} \). The expression for \( K_a \) is:

\[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]

Substituting the equilibrium concentrations into the expression:

\[ K_a = \frac{x \cdot (1 - x)}{x} \]

This simplifies to:

\[ K_a = 1 - x \]

### Step 4: Solve for \( x \)
Since \( K_a \) is very small compared to 1, we can approximate \( 1 - x \approx 1 \):

\[ 1.6 \times 10^{-5} = 1 - x \implies x \approx 1 - 1.6 \times 10^{-5} \approx 1.6 \times 10^{-5} \]

### Step 5: Calculate the concentration of \( H^+ \)
Since \( x \) represents the concentration of \( H^+ \) ions produced:

\[ [H^+] = x \approx 1.6 \times 10^{-5} \text{ M} \]

### Conclusion
Thus, the concentration of \( H^+ \) ions in the resulting solution is approximately \( 1.6 \times 10^{-5} \text{ M} \).

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