A solution of weak acid HA was titrated with base NaOH. The equivalent point was reached when 40 mL. Of 0.1 M NaOH has been added. Now 20 mL of 0.1 M HCl were added to titrated solution, the pH was found to be 5.0 What will be the pH of the solution obtained by mixing 20 mL of 0.2 M NaOH and 20 mL of 0.2 M HA?

Options

A. 7
B. 9
C. 11
D. None of these

Answer

To solve the problem step by step, we need to analyze the titration of the weak acid HA with NaOH and then determine the pH after mixing NaOH and HA.

### Step 1: Calculate the millimoles of NaOH at the equivalence point
At the equivalence point, the number of millimoles of NaOH added is calculated as follows:
\[ \text{Millimoles of NaOH} = \text{Volume (mL)} \times \text{Molarity (M)} = 40 \, \text{mL} \times 0.1 \, \text{M} = 4 \, \text{mmol} \]

**Hint:** Remember that at the equivalence point, the moles of acid and base are equal.

### Step 2: Determine the concentration of the salt formed (NaA)
From the reaction:
\[ \text{HA} + \text{NaOH} \rightarrow \text{NaA} + \text{H}_2\text{O} \]
At the equivalence point, all 4 mmol of HA has reacted with 4 mmol of NaOH to form 4 mmol of NaA.

### Step 3: Add HCl to the titrated solution
Now, we add 20 mL of 0.1 M HCl:
\[ \text{Millimoles of HCl} = 20 \, \text{mL} \times 0.1 \, \text{M} = 2 \, \text{mmol} \]
This HCl will react with the NaA:
\[ \text{NaA} + \text{HCl} \rightarrow \text{HA} + \text{NaCl} \]
Initially, we have 4 mmol of NaA and we add 2 mmol of HCl. After the reaction:
- Remaining NaA = \(4 \, \text{mmol} - 2 \, \text{mmol} = 2 \, \text{mmol}\)
- Formed HA = \(2 \, \text{mmol}\)

### Step 4: Calculate the pKa
Given that the pH after adding HCl is 5.0, we can use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
At this point:
- \([\text{A}^-] = 2 \, \text{mmol}\)
- \([\text{HA}] = 2 \, \text{mmol}\)

Substituting into the equation:
\[ 5.0 = \text{pKa} + \log\left(\frac{2}{2}\right) \]
\(\log(1) = 0\), thus:
\[ \text{pKa} = 5.0 \]

### Step 5: Mixing 20 mL of 0.2 M NaOH and 20 mL of 0.2 M HA
Now we consider the new mixture:
\[ \text{Millimoles of NaOH} = 20 \, \text{mL} \times 0.2 \, \text{M} = 4 \, \text{mmol} \]
\[ \text{Millimoles of HA} = 20 \, \text{mL} \times 0.2 \, \text{M} = 4 \, \text{mmol} \]
The reaction will proceed as:
\[ \text{HA} + \text{NaOH} \rightarrow \text{NaA} + \text{H}_2\text{O} \]
After the reaction, we will have:
- NaA = 4 mmol (since all HA and NaOH react completely)
- HA = 0 mmol (all HA is converted)

### Step 6: Calculate the concentration of NaA
The total volume after mixing is:
\[ 20 \, \text{mL} + 20 \, \text{mL} = 40 \, \text{mL} \]
The concentration of NaA is:
\[ [\text{NaA}] = \frac{4 \, \text{mmol}}{40 \, \text{mL}} = 0.1 \, \text{M} \]

### Step 7: Calculate the pH using the Henderson-Hasselbalch equation
Using the Henderson-Hasselbalch equation again:
\[ \text{pH} = \frac{1}{2} \text{pK}_w + \text{pKa} + \log C \]
Where \(C = 0.1 \, \text{M}\):
\[ \text{pK}_w = 14 \Rightarrow \frac{1}{2} \text{pK}_w = 7 \]
Substituting the values:
\[ \text{pH} = 7 + 5 + \log(0.1) = 7 + 5 - 1 = 11 \]

### Final Answer
The pH of the solution obtained by mixing 20 mL of 0.2 M NaOH and 20 mL of 0.2 M HA is **11**.

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