`"0.5 g NaOH"` was added to `"200 ml 0.1 M HCl"`, final concentration of reactant left is :

Options

A. `3//80`
B. `2//80`
C. `3//40`
D. `5//80`

Answer

To find the final concentration of the reactant left after the reaction between NaOH and HCl, we will follow these steps:

### Step 1: Calculate the number of moles of NaOH
Given:
- Mass of NaOH = 0.5 g
- Molar mass of NaOH = 40 g/mol

To find the number of moles of NaOH:
\[ \text{Number of moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{0.5 \, \text{g}}{40 \, \text{g/mol}} = 0.0125 \, \text{mol} \]

### Step 2: Calculate the number of moles of HCl
Given:
- Volume of HCl solution = 200 ml = 0.2 L
- Molarity of HCl = 0.1 M

To find the number of moles of HCl:
\[ \text{Number of moles of HCl} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.2 \, \text{L} = 0.02 \, \text{mol} \]

### Step 3: Determine the limiting reactant
The balanced chemical reaction is:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

From the stoichiometry of the reaction, 1 mole of NaOH reacts with 1 mole of HCl.

- Moles of NaOH = 0.0125 mol
- Moles of HCl = 0.02 mol

Since NaOH is present in lesser amount (0.0125 mol), it is the limiting reactant.

### Step 4: Calculate the moles of HCl that react
Since NaOH is the limiting reactant, all of it will react with an equivalent amount of HCl:
\[ \text{Moles of HCl that react} = \text{Moles of NaOH} = 0.0125 \, \text{mol} \]

### Step 5: Calculate the remaining moles of HCl
\[ \text{Remaining moles of HCl} = \text{Initial moles of HCl} - \text{Moles of HCl that react} = 0.02 \, \text{mol} - 0.0125 \, \text{mol} = 0.0075 \, \text{mol} \]

### Step 6: Calculate the final concentration of remaining HCl
The total volume of the solution after the reaction remains approximately the same (200 ml = 0.2 L).

To find the final concentration of the remaining HCl:
\[ \text{Final concentration of HCl} = \frac{\text{Remaining moles of HCl}}{\text{Total volume in L}} = \frac{0.0075 \, \text{mol}}{0.2 \, \text{L}} = 0.0375 \, \text{M} \]

### Final Answer
The final concentration of the reactant left (HCl) is **0.0375 M**.

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