Four grams of NaOH solid are dissolved in just enough water to make I litre of solution. What is the `[H^(+)]` . of the solution ?

Options

A. `10^(-2)` moles /litre
B. `10^(-1)` moles /litre
C. `10^(-12) ` moles /litre
D. `10^(-13) ` moles /litre

Answer

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

### Step 1: Calculate the molarity of NaOH
First, we need to determine the number of moles of NaOH in 4 grams.

1. **Find the molecular weight of NaOH:**
- Sodium (Na) = 23 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol
- Molecular weight of NaOH = 23 + 16 + 1 = 40 g/mol

2. **Calculate the number of moles of NaOH:**
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molecular weight}} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ moles} \]

3. **Calculate the molarity of the solution:**
Since the solution is made to 1 liter, the molarity (M) is:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.1 \text{ moles}}{1 \text{ L}} = 0.1 \text{ M} \]

### Step 2: Determine the concentration of hydroxide ions \([OH^-]\)
Since NaOH is a strong base, it completely dissociates in water:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Thus, the concentration of hydroxide ions \([OH^-]\) is equal to the molarity of NaOH:
\[ [OH^-] = 0.1 \text{ M} \]

### Step 3: Use the ion product of water to find \([H^+]\)
The ion product of water at 25°C is:
\[ K_w = [H^+][OH^-] = 1 \times 10^{-14} \]
We can rearrange this equation to find \([H^+]\):
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{0.1} \]
Calculating this gives:
\[ [H^+] = 1 \times 10^{-13} \text{ M} \]

### Final Answer
The concentration of hydrogen ions \([H^+]\) in the solution is:
\[ [H^+] = 1 \times 10^{-13} \text{ M} \]