If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M `HCOOH [K_a= 2 xx 10^(-4)].` The pOH of the resulting solution is

Options

A. 3.4
B. 3.7
C. 7
D. 10.3

Answer

To solve the problem, we need to find the pOH of the resulting solution after mixing 40 ml of 0.2 M KOH with 160 ml of 0.1 M HCOOH. Here’s a step-by-step solution:

### Step 1: Calculate the milliequivalents of KOH and HCOOH
- **KOH**:
\[ \text{Milliequivalents of KOH} = \text{Volume (ml)} \times \text{Molarity (M)} = 40 \, \text{ml} \times 0.2 \, \text{M} = 8 \, \text{meq} \]

- **HCOOH**:
\[ \text{Milliequivalents of HCOOH} = \text{Volume (ml)} \times \text{Molarity (M)} = 160 \, \text{ml} \times 0.1 \, \text{M} = 16 \, \text{meq} \]

### Step 2: Determine the reaction between KOH and HCOOH
KOH will react with HCOOH as follows:
\[ \text{HCOOH} + \text{KOH} \rightarrow \text{HCOOK} + \text{H}_2\text{O} \]
- KOH (8 meq) will react with HCOOH (8 meq), producing 8 meq of HCOOK (potassium formate).
- Remaining HCOOH after reaction:
\[ 16 \, \text{meq} - 8 \, \text{meq} = 8 \, \text{meq} \text{ of HCOOH} \]

### Step 3: Calculate the total volume of the solution
\[ \text{Total Volume} = 40 \, \text{ml} + 160 \, \text{ml} = 200 \, \text{ml} \]

### Step 4: Calculate concentrations of HCOOK and HCOOH
- **Concentration of HCOOK**:
\[ \text{Concentration of HCOOK} = \frac{8 \, \text{meq}}{200 \, \text{ml}} = 0.04 \, \text{M} \]

- **Concentration of HCOOH**:
\[ \text{Concentration of HCOOH} = \frac{8 \, \text{meq}}{200 \, \text{ml}} = 0.04 \, \text{M} \]

### Step 5: Calculate pKa from Ka
Given \( K_a = 2 \times 10^{-4} \):
\[ pK_a = -\log(K_a) = -\log(2 \times 10^{-4}) \approx 3.7 \]

### Step 6: Use the Henderson-Hasselbalch equation to find pH
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Since the concentrations of HCOOK and HCOOH are equal:
\[ pH = 3.7 + \log\left(\frac{0.04}{0.04}\right) = 3.7 + 0 = 3.7 \]

### Step 7: Calculate pOH
Using the relationship \( pOH + pH = 14 \):
\[ pOH = 14 - pH = 14 - 3.7 = 10.3 \]

### Final Answer
The pOH of the resulting solution is **10.3**.

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