If `50ml` of `0.2 M KOH` is added to `40 ml` of `0.05 M HCOOH`, the `pH` of the resulting solution is (`K_(a)=1.8xx10^(-4)`)

Options

A. `3.4`
B. `7.5`
C. `5.6`
D. `3.75`

Explanation

To solve the problem of finding the pH of the resulting solution when `50 ml` of `0.2 M KOH` is added to `40 ml` of `0.05 M HCOOH`, we can follow these steps:

### Step 1: Calculate the moles of KOH and HCOOH
First, we need to calculate the number of moles of KOH and HCOOH present in their respective solutions.

- **Moles of KOH**:
\[ \text{Moles of KOH} = \text{Concentration} \times \text{Volume} = 0.2 \, \text{M} \times 0.050 \, \text{L} = 0.010 \, \text{mol} \]

- **Moles of HCOOH**:
\[ \text{Moles of HCOOH} = \text{Concentration} \times \text{Volume} = 0.05 \, \text{M} \times 0.040 \, \text{L} = 0.002 \, \text{mol} \]

### Step 2: Determine the limiting reactant and the reaction
KOH is a strong base and HCOOH is a weak acid. The reaction between them can be represented as:
\[ \text{HCOOH} + \text{KOH} \rightarrow \text{HCOOK} + \text{H}_2\text{O} \]

From the calculations:
- Moles of KOH = 0.010 mol
- Moles of HCOOH = 0.002 mol

Since HCOOH is the limiting reactant, it will completely react with KOH.

### Step 3: Calculate the remaining moles after the reaction
- Moles of KOH remaining after reaction:
\[ \text{Remaining KOH} = 0.010 \, \text{mol} - 0.002 \, \text{mol} = 0.008 \, \text{mol} \]
- Moles of HCOOH remaining after reaction:
\[ \text{Remaining HCOOH} = 0.002 \, \text{mol} - 0.002 \, \text{mol} = 0 \, \text{mol} \]
- Moles of salt (HCOOK) produced:
\[ \text{Moles of HCOOK} = 0.002 \, \text{mol} \]

### Step 4: Calculate the total volume of the solution
The total volume after mixing:
\[ \text{Total Volume} = 50 \, \text{ml} + 40 \, \text{ml} = 90 \, \text{ml} = 0.090 \, \text{L} \]

### Step 5: Calculate the concentrations of KOH and HCOOK
- **Concentration of KOH**:
\[ \text{Concentration of KOH} = \frac{0.008 \, \text{mol}}{0.090 \, \text{L}} \approx 0.0889 \, \text{M} \]

- **Concentration of HCOOK**:
\[ \text{Concentration of HCOOK} = \frac{0.002 \, \text{mol}}{0.090 \, \text{L}} \approx 0.0222 \, \text{M} \]

### Step 6: Calculate the pKa of HCOOH
Given \( K_a = 1.8 \times 10^{-4} \):
\[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-4}) \approx 3.74 \]

### Step 7: Use the Henderson-Hasselbalch equation
Since we have a strong base (KOH) and a weak acid (HCOOH), we can use the Henderson-Hasselbalch equation for the buffer solution formed:
\[ pH = pK_a + \log\left(\frac{[Salt]}{[Acid]}\right) \]
In this case:
\[ pH = pK_a + \log\left(\frac{0.0222}{0}\right) \]
Since the weak acid is completely neutralized, we consider the remaining KOH concentration.

### Step 8: Calculate the pH
Since KOH is in excess, we can approximate the pH from the concentration of KOH:
\[ pOH = -\log(0.0889) \approx 1.05 \]
\[ pH = 14 - pOH \approx 14 - 1.05 = 12.95 \]

### Final Answer
Thus, the pH of the resulting solution is approximately **12.95**.

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