To find the concentration of \( H^+ \) ions in the solution formed by mixing 0.5 moles of HCl and 0.5 moles of sodium acetate (CH₃COONa) in 500 mL of water, we can follow these steps:
### Step 1: Determine the concentrations of HCl and CH₃COONa
- We have 0.5 moles of HCl and 0.5 moles of CH₃COONa dissolved in 500 mL of solution.
- To find the concentration (C) in mol/L, we convert 500 mL to liters:
\[
500 \, \text{mL} = 0.5 \, \text{L}
\]
- Now, calculate the concentration:
\[
C_{\text{HCl}} = \frac{0.5 \, \text{moles}}{0.5 \, \text{L}} = 1 \, \text{M}
\]
\[
C_{\text{CH}_3\text{COONa}} = \frac{0.5 \, \text{moles}}{0.5 \, \text{L}} = 1 \, \text{M}
\]
### Step 2: Write the reaction
- HCl is a strong acid and will completely dissociate in solution:
\[
\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-
\]
- Sodium acetate (CH₃COONa) will dissociate into acetate ions (CH₃COO⁻) and sodium ions (Na⁺):
\[
\text{CH}_3\text{COONa} \rightarrow \text{CH}_3\text{COO}^- + \text{Na}^+
\]
- The acetate ion can react with the \( H^+ \) ions:
\[
\text{CH}_3\text{COO}^- + \text{H}^+ \rightleftharpoons \text{CH}_3\text{COOH}
\]
### Step 3: Set up the equilibrium expression
- Let \( C \) be the initial concentration of both HCl and CH₃COONa, which is 1 M.
- At equilibrium, let \( \alpha \) be the degree of dissociation of acetic acid (CH₃COOH).
- The concentrations at equilibrium will be:
- \( [\text{CH}_3\text{COOH}] = C \alpha \)
- \( [\text{H}^+] = C - C \alpha = C(1 - \alpha) \)
- \( [\text{CH}_3\text{COO}^-] = C - C \alpha = C(1 - \alpha) \)
### Step 4: Apply the acid dissociation constant (Ka)
- The dissociation constant \( K_a \) for acetic acid is given as \( 1.6 \times 10^{-5} \):
\[
K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}
\]
- Substituting the equilibrium concentrations into the expression:
\[
1.6 \times 10^{-5} = \frac{(C(1 - \alpha))(C(1 - \alpha))}{C \alpha}
\]
\[
1.6 \times 10^{-5} = \frac{C^2(1 - \alpha)^2}{C \alpha}
\]
\[
1.6 \times 10^{-5} = \frac{C(1 - \alpha)^2}{\alpha}
\]
### Step 5: Substitute \( C \) and simplify
- Since \( C = 1 \, \text{M} \):
\[
1.6 \times 10^{-5} = \frac{(1)(1 - \alpha)^2}{\alpha}
\]
- Rearranging gives:
\[
1.6 \times 10^{-5} \alpha = (1 - \alpha)^2
\]
### Step 6: Assume \( \alpha \) is small
- Since \( K_a \) is small, we can assume \( \alpha \) is much smaller than 1, thus \( 1 - \alpha \approx 1 \):
\[
1.6 \times 10^{-5} \alpha \approx 1
\]
\[
\alpha^2 \approx 1.6 \times 10^{-5}
\]
\[
\alpha \approx \sqrt{1.6 \times 10^{-5}} \approx 0.004
\]
### Step 7: Calculate \( [H^+] \)
- The concentration of \( H^+ \) ions is:
\[
[H^+] = C \alpha = 1 \times 0.004 = 0.004 \, \text{M} = 4 \times 10^{-3} \, \text{M}
\]
### Final Answer
The concentration of \( H^+ \) ions in the resulting solution is \( 4 \times 10^{-3} \, \text{M} \).
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