What will be the normality of the salt solution obtained by neutralizing x mL of y (N) HCl with y mL of x (N) NaOH and finally adding (x + y) mL distilled water ?

Options

A. `(2(x+y))/(xy)N`
B. `(xy)/(2(x+y))N`
C. `(2xy)/(x+y)N`
D. `((x+y)/(xy))N`

Explanation

To find the normality of the salt solution obtained by neutralizing \( x \) mL of \( y \) N HCl with \( y \) mL of \( x \) N NaOH and then adding \( (x + y) \) mL of distilled water, we can follow these steps:

### Step 1: Calculate the equivalents of HCl
The number of equivalents of HCl can be calculated using the formula:
\[ \text{Equivalents of HCl} = \text{Normality} \times \text{Volume (L)} \]
Given that the normality of HCl is \( y \) N and the volume is \( x \) mL (which is \( \frac{x}{1000} \) L), we have:
\[ \text{Equivalents of HCl} = y \times \frac{x}{1000} = \frac{xy}{1000} \]

### Step 2: Calculate the equivalents of NaOH
Similarly, the equivalents of NaOH can be calculated using the same formula:
\[ \text{Equivalents of NaOH} = \text{Normality} \times \text{Volume (L)} \]
Given that the normality of NaOH is \( x \) N and the volume is \( y \) mL (which is \( \frac{y}{1000} \) L), we have:
\[ \text{Equivalents of NaOH} = x \times \frac{y}{1000} = \frac{xy}{1000} \]

### Step 3: Determine the amount of salt formed
In the neutralization reaction:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
The stoichiometry shows that one equivalent of HCl reacts with one equivalent of NaOH to produce one equivalent of NaCl. Thus, the number of equivalents of NaCl formed will be equal to the number of equivalents of HCl or NaOH, which is:
\[ \text{Equivalents of NaCl} = \frac{xy}{1000} \]

### Step 4: Calculate the total volume of the solution
After the reaction, we add \( (x + y) \) mL of distilled water to the solution. The total volume of the solution is:
\[ \text{Total Volume} = x + y + (x + y) = 2(x + y) \text{ mL} \]
Converting this to liters, we have:
\[ \text{Total Volume (L)} = \frac{2(x + y)}{1000} \]

### Step 5: Calculate the normality of the salt solution
Normality is defined as the number of equivalents of solute per liter of solution. Therefore, the normality of the NaCl solution can be calculated as:
\[ \text{Normality} = \frac{\text{Equivalents of NaCl}}{\text{Total Volume (L)}} \]
Substituting the values we calculated:
\[ \text{Normality} = \frac{\frac{xy}{1000}}{\frac{2(x + y)}{1000}} = \frac{xy}{2(x + y)} \]

### Final Answer
Thus, the normality of the salt solution is:
\[ \text{Normality} = \frac{xy}{2(x + y)} \]

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