`[H^(+)]` conc. Of 4% NaOH solution is

Options

A. 1m ionn litre
B. `6 xx 10^(23)`
C. `6 xx 10^(9)`
D. `10^(-14)` gm ion/litre

Answer

To find the concentration of \([H^+]\) in a 4% NaOH solution, we can follow these steps:

### Step 1: Calculate the mass of NaOH in the solution
A 4% NaOH solution means that there are 4 grams of NaOH in 100 mL of solution.

### Step 2: Calculate the molar mass of NaOH
The molar mass of NaOH can be calculated as follows:
- Sodium (Na) = 23 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol

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

### Step 3: Calculate the number of moles of NaOH
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 4: Calculate the molarity of NaOH
Molarity (M) is defined as the number of moles of solute per liter of solution. Since we have 100 mL of solution, we convert this to liters:
\[ \text{Volume in liters} = \frac{100 \text{ mL}}{1000} = 0.1 \text{ L} \]
Now, we can calculate the molarity:
\[ \text{Molarity} = \frac{0.1 \text{ moles}}{0.1 \text{ L}} = 1 \text{ M} \]

### Step 5: Determine the concentration of \([OH^-]\)
In a NaOH solution, NaOH completely dissociates into Na\(^+\) and OH\(^-\) ions. Thus, the concentration of \([OH^-]\) is equal to the molarity of the NaOH solution:
\[ [OH^-] = 1 \text{ M} \]

### Step 6: Use the ionic product of water to find \([H^+]\)
The ionic product of water at 25°C is given by:
\[ K_w = [H^+][OH^-] = 10^{-14} \]
We know \([OH^-] = 1 \text{ M}\). Therefore, we can find \([H^+]\):
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{1} = 10^{-14} \text{ M} \]

### Final Answer
The concentration of \([H^+]\) in a 4% NaOH solution is:
\[ [H^+] = 10^{-14} \text{ M} \]

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