`[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 hydrogen ions \([H^+]\) in a 4% NaOH solution, we can follow these steps:

### Step 1: Determine the number of moles of NaOH
Given that the solution is 4% NaOH by weight, this means there are 4 grams of NaOH in 100 mL of solution.

To find the number of moles of NaOH, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of NaOH is approximately 40 g/mol. Thus:
\[ \text{Number of moles of NaOH} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 2: 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 need to convert this to liters:
\[ 100 \text{ mL} = 0.1 \text{ L} \]
Now, we can calculate the molarity:
\[ \text{Molarity of NaOH} = \frac{0.1 \text{ moles}}{0.1 \text{ L}} = 1 \text{ M} \]

### Step 3: Determine the concentration of hydroxide ions \([OH^-]\)
Since NaOH is a strong base, it dissociates completely in water:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
This means that the concentration of hydroxide ions \([OH^-]\) is equal to the molarity of NaOH:
\[ [OH^-] = 1 \text{ M} \]

### Step 4: Use the ionic product of water to find \([H^+]\)
At 25°C, the ionic product of water (\(K_w\)) is:
\[ K_w = [H^+][OH^-] = 1 \times 10^{-14} \]
We can rearrange this equation to find \([H^+]\):
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{1} = 1 \times 10^{-14} \text{ M} \]

### Final Answer
The concentration of hydrogen ions \([H^+]\) in a 4% NaOH solution is:
\[ [H^+] = 1 \times 10^{-14} \text{ M} \]

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