A 50 ml sample of acetic acid was titrated with 0.012 M KOH and 38.62 ml of base were required to reach equivalence point. What was the pH of the titration mixture when 19.31 ml of base had been added? `[pK_(a)` (acetic acid) = 4.74]

Options

A. 2.94
B. 3.54
C. 4.74
D. 5.74

Answer

To solve the problem step-by-step, we will follow these steps:

### Step 1: Identify the components involved in the titration
In this titration, we have acetic acid (CH₃COOH) being titrated with potassium hydroxide (KOH). At the equivalence point, all the acetic acid has been converted into its salt, potassium acetate (CH₃COOK).

### Step 2: Calculate the moles of acetic acid
Given that the volume of acetic acid is 50 mL, we need to find the number of moles of acetic acid. However, the concentration of acetic acid is not provided, so we will denote it as \(C_a\) (in mol/L).

### Step 3: Calculate the moles of KOH used at the equivalence point
At the equivalence point, we know that 38.62 mL of KOH is used. The concentration of KOH is given as 0.012 M.

\[ \text{Moles of KOH} = \text{Volume (L)} \times \text{Concentration (mol/L)} = 0.03862 \, \text{L} \times 0.012 \, \text{mol/L} = 0.00046344 \, \text{mol} \]

### Step 4: Determine the moles of acetic acid
At the equivalence point, the moles of KOH will equal the moles of acetic acid. Therefore, the moles of acetic acid is also 0.00046344 mol.

### Step 5: Calculate the concentration of acetic acid
The concentration of acetic acid can be calculated as follows:

\[ C_a = \frac{\text{Moles of acetic acid}}{\text{Volume of acetic acid (L)}} = \frac{0.00046344 \, \text{mol}}{0.050 \, \text{L}} = 0.0092688 \, \text{M} \]

### Step 6: Calculate the moles of KOH added at 19.31 mL
Now, we need to find the moles of KOH added when 19.31 mL is used:

\[ \text{Moles of KOH added} = 0.01931 \, \text{L} \times 0.012 \, \text{mol/L} = 0.00023172 \, \text{mol} \]

### Step 7: Determine the remaining acetic acid and formed acetate
At this point, we have added 0.00023172 mol of KOH. The moles of acetic acid remaining can be calculated as follows:

\[ \text{Remaining acetic acid} = \text{Initial moles of acetic acid} - \text{Moles of KOH added} = 0.00046344 \, \text{mol} - 0.00023172 \, \text{mol} = 0.00023172 \, \text{mol} \]

The moles of acetate formed (from KOH reacting with acetic acid) is equal to the moles of KOH added:

\[ \text{Moles of acetate (CH₃COOK)} = 0.00023172 \, \text{mol} \]

### Step 8: Calculate the concentrations of acetic acid and acetate
Now we can calculate the concentrations of acetic acid and acetate in the total volume of the solution, which is \(50 \, \text{mL} + 19.31 \, \text{mL} = 69.31 \, \text{mL} = 0.06931 \, \text{L}\).

\[ [\text{CH}_3\text{COOH}] = \frac{0.00023172 \, \text{mol}}{0.06931 \, \text{L}} = 0.003344 \, \text{M} \]

\[ [\text{CH}_3\text{COOK}] = \frac{0.00023172 \, \text{mol}}{0.06931 \, \text{L}} = 0.003344 \, \text{M} \]

### Step 9: Use the Henderson-Hasselbalch equation
Since we have a buffer solution, we can use the Henderson-Hasselbalch equation to find the pH:

\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) \]

Here, \([\text{base}] = [\text{CH}_3\text{COOK}]\) and \([\text{acid}] = [\text{CH}_3\text{COOH}]\). Since both concentrations are equal, the log term becomes zero:

\[ \text{pH} = \text{pK}_a + \log(1) = \text{pK}_a \]

Given that \(pK_a = 4.74\):

\[ \text{pH} = 4.74 \]

### Final Answer
The pH of the titration mixture when 19.31 mL of KOH has been added is **4.74**.

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