`"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 solve the problem of determining the final concentration of the reactant left after adding 0.5 g of NaOH to 200 ml of 0.1 M HCl, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) can be represented as:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 2: Calculate the moles of NaOH
To find the number of moles of NaOH, we use the formula:
\[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of NaOH is calculated as follows:
- Na: 23 g/mol
- O: 16 g/mol
- H: 1 g/mol

Thus, the molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.

Now, calculate the moles of NaOH:
\[ \text{Moles of NaOH} = \frac{0.5 \text{ g}}{40 \text{ g/mol}} = 0.0125 \text{ moles} \]

### Step 3: Calculate the moles of HCl
Using the molarity and volume of HCl, we can find the moles:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \]

Given that the volume of HCl is 200 ml (which is 0.2 L) and the molarity is 0.1 M:
\[ \text{Moles of HCl} = 0.1 \text{ M} \times 0.2 \text{ L} = 0.02 \text{ moles} \]

### Step 4: Identify the limiting reagent
From the balanced equation, we see that NaOH and HCl react in a 1:1 ratio. We have:
- Moles of NaOH = 0.0125
- Moles of HCl = 0.02

Since we have less NaOH than HCl, NaOH is the limiting reagent.

### Step 5: Calculate the remaining moles of HCl after the reaction
Since NaOH is the limiting reagent, it will be completely consumed. The moles of HCl that will react with NaOH are also 0.0125 moles (1:1 ratio).

Remaining moles of HCl:
\[ \text{Remaining HCl} = \text{Initial HCl} - \text{Reacted HCl} = 0.02 \text{ moles} - 0.0125 \text{ moles} = 0.0075 \text{ moles} \]

### Step 6: Calculate the final concentration of the remaining HCl
To find the concentration (molarity) of the remaining HCl, we use the formula:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \]

The volume of the solution remains 200 ml (0.2 L):
\[ \text{Molarity of remaining HCl} = \frac{0.0075 \text{ moles}}{0.2 \text{ L}} = 0.0375 \text{ M} \]

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

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