Calculate the hydrogen ion concentration (`"in mol"//dm^3`) in a solution containing 0.04 mole of acetic acid and 0.05 mole of sodium acetate in 500 mL of solution.Dissociation constant for acetic acid is `1.75xx10^(-6)` Report your answer after multiplying by `2xx10^(6)`

Explanation

To calculate the hydrogen ion concentration in a solution containing acetic acid and sodium acetate, we can follow these steps:

### Step 1: Identify the components of the solution
We have:
- Acetic acid (weak acid) = 0.04 moles
- Sodium acetate (conjugate base) = 0.05 moles
- Volume of the solution = 500 mL = 0.5 L

### Step 2: Calculate the concentrations of acetic acid and sodium acetate
To find the concentrations, we use the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in L}} \]

For acetic acid:
\[ [\text{Acetic Acid}] = \frac{0.04 \text{ moles}}{0.5 \text{ L}} = 0.08 \text{ mol/dm}^3 \]

For sodium acetate:
\[ [\text{Sodium Acetate}] = \frac{0.05 \text{ moles}}{0.5 \text{ L}} = 0.10 \text{ mol/dm}^3 \]

### Step 3: Calculate the pKa from the dissociation constant (Ka)
We are given the dissociation constant for acetic acid:
\[ K_a = 1.75 \times 10^{-6} \]
To find pKa:
\[ pK_a = -\log(K_a) = -\log(1.75 \times 10^{-6}) \approx 5.75 \]

### Step 4: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for a buffer solution is:
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

Substituting the values:
\[ pH = 5.75 + \log\left(\frac{0.10}{0.08}\right) \]
Calculating the log term:
\[ \log\left(\frac{0.10}{0.08}\right) = \log(1.25) \approx 0.0969 \]
Now substituting back:
\[ pH = 5.75 + 0.0969 \approx 5.8469 \]

### Step 5: Calculate the hydrogen ion concentration
Using the relationship between pH and hydrogen ion concentration:
\[ \text{pH} = -\log[\text{H}^+] \]
This implies:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-5.8469} \]
Calculating this gives:
\[ [\text{H}^+] \approx 1.4 \times 10^{-6} \text{ mol/dm}^3 \]

### Step 6: Multiply the hydrogen ion concentration by \(2 \times 10^6\)
Finally, we need to report the answer after multiplying by \(2 \times 10^6\):
\[ \text{Final concentration} = 1.4 \times 10^{-6} \times 2 \times 10^6 = 2.8 \]

### Final Answer
The hydrogen ion concentration is \(2.8\).

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