calculate the amount of `KOH` requried to neutralise 15 mEq of the following:
a. `HCl`

Explanation

To calculate the amount of KOH required to neutralize 15 mEq of HCl, follow these steps:

### Step 1: Understand the Reaction
The neutralization reaction between KOH and HCl can be represented as:
\[ \text{KOH} + \text{HCl} \rightarrow \text{KCl} + \text{H}_2\text{O} \]

### Step 2: Determine the Equivalence
In acid-base reactions, one equivalent of a base neutralizes one equivalent of an acid. Therefore, to neutralize 15 mEq of HCl, you will need 15 mEq of KOH.

### Step 3: Convert Milliequivalents to Equivalents
1 mEq is equal to \(10^{-3}\) equivalents. Thus, 15 mEq can be converted to equivalents:
\[ 15 \text{ mEq} = 15 \times 10^{-3} \text{ equivalents} \]

### Step 4: Find the Equivalent Weight of KOH
The equivalent weight of KOH is calculated as follows:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Valency}} \]
The molar mass of KOH is approximately 56.1 g/mol, and since KOH provides one OH\(^-\) ion, its valency factor is 1. Thus, the equivalent weight of KOH is:
\[ \text{Equivalent Weight of KOH} = 56.1 \text{ g/mol} \]

### Step 5: Calculate the Amount of KOH Required
To find the amount of KOH required in grams, use the formula:
\[ \text{Weight (g)} = \text{Equivalent Weight (g/equiv)} \times \text{Number of Equivalents} \]
Substituting the values:
\[ \text{Weight of KOH} = 56.1 \text{ g/equiv} \times 15 \times 10^{-3} \text{ equiv} \]
\[ \text{Weight of KOH} = 56.1 \times 0.015 \]
\[ \text{Weight of KOH} = 0.8415 \text{ g} \]

### Conclusion
The amount of KOH required to neutralize 15 mEq of HCl is approximately **0.8415 grams**.

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