Calculate the pH of solution prepared by mixing 50.0 mL of 0.200 M `HC_2H_3O_2` and 50.0 mL of 0.100 M NaOH. `[Ka(CH_3COOH)=1.8 xx 10^(-5)]`

Answer

To calculate the pH of the solution prepared by mixing 50.0 mL of 0.200 M acetic acid (HC₂H₃O₂) and 50.0 mL of 0.100 M sodium hydroxide (NaOH), we will follow these steps:

### Step 1: Calculate the number of millimoles of acetic acid and sodium hydroxide.
1. **For acetic acid (HC₂H₃O₂)**:
- Molarity = 0.200 M
- Volume = 50.0 mL
- Millimoles of acetic acid = Molarity × Volume (in mL) = 0.200 mol/L × 50.0 mL = 10.0 mmol

2. **For sodium hydroxide (NaOH)**:
- Molarity = 0.100 M
- Volume = 50.0 mL
- Millimoles of sodium hydroxide = Molarity × Volume (in mL) = 0.100 mol/L × 50.0 mL = 5.0 mmol

### Step 2: Determine the limiting reagent.
- The stoichiometry of the reaction between acetic acid and sodium hydroxide is 1:1.
- We have 10.0 mmol of acetic acid and 5.0 mmol of sodium hydroxide.
- Since sodium hydroxide is present in a lesser amount (5.0 mmol), it is the limiting reagent.

### Step 3: Calculate the remaining acetic acid and the amount of sodium acetate formed.
- Sodium hydroxide will react completely with 5.0 mmol of acetic acid.
- Remaining acetic acid = Initial acetic acid - Reacted acetic acid
- Remaining acetic acid = 10.0 mmol - 5.0 mmol = 5.0 mmol
- Sodium acetate (CH₃COONa) is formed in a 1:1 ratio with sodium hydroxide.
- Therefore, the amount of sodium acetate formed = 5.0 mmol.

### Step 4: Calculate the concentrations of acetic acid and sodium acetate in the final solution.
- The total volume of the solution after mixing = 50.0 mL + 50.0 mL = 100.0 mL = 0.100 L.
- Concentration of acetic acid = Remaining acetic acid / Total volume = 5.0 mmol / 100.0 mL = 0.050 M.
- Concentration of sodium acetate = Sodium acetate / Total volume = 5.0 mmol / 100.0 mL = 0.050 M.

### Step 5: Use the Henderson-Hasselbalch equation to find the pH.
- The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
where:
- \([\text{A}^-]\) is the concentration of the salt (sodium acetate),
- \([\text{HA}]\) is the concentration of the weak acid (acetic acid).

- Given:
- \(K_a\) for acetic acid = \(1.8 \times 10^{-5}\)
- Calculate \(pK_a\):
\[ pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

- Substitute the values into the Henderson-Hasselbalch equation:
\[ \text{pH} = 4.74 + \log\left(\frac{0.050}{0.050}\right) \]
\[ \text{pH} = 4.74 + \log(1) = 4.74 + 0 = 4.74 \]

### Final Answer:
The pH of the solution is **4.74**.

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