A solution of KCl and KOH was neutralized with 120 ml of 0.12 (N) HCl. What is the amount of KOH in the mixture -

Options

A. 1.5gm
B. 0.806gm
C. 1.2 gm
D. 0.88gm

Answer

To find the amount of KOH in the mixture that was neutralized by HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Volume of HCl solution (V) = 120 ml
- Normality of HCl (N) = 0.12 N

2. **Calculate the Milliequivalents of HCl:**
The formula for milliequivalents (mEq) is:
\[ \text{mEq} = \text{Normality} \times \text{Volume (ml)} \]
Substituting the values:
\[ \text{mEq of HCl} = 0.12 \, \text{N} \times 120 \, \text{ml} = 14.4 \, \text{mEq} \]

3. **Understand the Reaction:**
The reaction between KOH and HCl is:
\[ \text{KOH} + \text{HCl} \rightarrow \text{KCl} + \text{H}_2\text{O} \]
Here, 1 mole of KOH reacts with 1 mole of HCl. Therefore, the milliequivalents of KOH will be equal to the milliequivalents of HCl.

4. **Calculate the Milliequivalents of KOH:**
Since the milliequivalents of KOH is equal to that of HCl:
\[ \text{mEq of KOH} = 14.4 \, \text{mEq} \]

5. **Find the Equivalent Weight of KOH:**
The equivalent weight (EW) of KOH can be calculated using its molecular weight (MW) and the n-factor (which is 1 for KOH):
\[ \text{Molecular weight of KOH} = 39 \, (\text{K}) + 16 \, (\text{O}) + 1 \, (\text{H}) = 56 \, \text{g/mol} \]
Therefore, the equivalent weight of KOH is:
\[ \text{EW of KOH} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{56}{1} = 56 \, \text{g/equiv} \]

6. **Calculate the Weight of KOH:**
The weight of KOH can be calculated using the formula:
\[ \text{Weight of KOH} = \frac{\text{mEq of KOH} \times \text{EW of KOH}}{1000} \]
Substituting the values:
\[ \text{Weight of KOH} = \frac{14.4 \, \text{mEq} \times 56 \, \text{g/equiv}}{1000} = 0.8064 \, \text{g} \]

7. **Final Result:**
The amount of KOH in the mixture is approximately **0.806 g**.