What will be the `H^+` con when 4 gm NaOH dissolved in 1000 ml. of water

Options

A. `10^(-1)`
B. `10^(-13)`
C. `10^(-4)`
D. `10^(-10)`

Answer

To solve the problem of finding the concentration of \( H^+ \) ions when 4 g of NaOH is dissolved in 1000 ml of water, we can follow these steps:

### Step 1: Calculate the number of moles of NaOH

We know that the number of moles can be calculated using the formula:

\[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \]

The molar mass of NaOH (sodium hydroxide) is approximately 40 g/mol.

Given:
- Mass of NaOH = 4 g
- Molar Mass of NaOH = 40 g/mol

Now, substituting the values:

\[ \text{Moles of NaOH} = \frac{4 \, \text{g}}{40 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 2: Calculate the molarity of NaOH solution

Molarity (M) is defined as the number of moles of solute per liter of solution. Since the volume of the solution is given in milliliters, we need to convert it to liters:

\[ \text{Volume} = 1000 \, \text{ml} = 1 \, \text{L} \]

Now, we can calculate the molarity:

\[ \text{Molarity} = \frac{\text{Moles of NaOH}}{\text{Volume of solution (L)}} = \frac{0.1 \, \text{mol}}{1 \, \text{L}} = 0.1 \, \text{M} \]

### Step 3: Determine the concentration of \( OH^- \) ions

Since NaOH completely dissociates in water, the concentration of \( OH^- \) ions will be equal to the molarity of NaOH:

\[ [OH^-] = 0.1 \, \text{M} \]

### Step 4: Use the ion product of water to find \( H^+ \) concentration

The ion product of water at 25°C is given by:

\[ K_w = [H^+][OH^-] = 1 \times 10^{-14} \]

We can rearrange this equation to find the concentration of \( H^+ \):

\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{0.1} = 1 \times 10^{-13} \, \text{M} \]

### Conclusion

The concentration of \( H^+ \) ions in the solution is:

\[ [H^+] = 1 \times 10^{-13} \, \text{M} \]