A monobasic weak acid solution which is 0.002 M has pH value equal to 5, The percentage ionization value of the acid in the solution will be

Options

A. 0.5
B. 0.005
C. 5
D. 0.05

Answer

To solve the problem of finding the percentage ionization of a monobasic weak acid solution with a concentration of 0.002 M and a pH of 5, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Concentration of the weak acid (C) = 0.002 M
- pH = 5

2. **Calculate the concentration of hydrogen ions ([H⁺]):**
- We know that pH = -log[H⁺].
- Therefore, [H⁺] = 10^(-pH) = 10^(-5) M.

3. **Set up the ionization equation:**
- For a monobasic weak acid (HX), the ionization can be represented as:
\[ HX \rightleftharpoons H^+ + X^- \]
- Initially, the concentration of HX is C (0.002 M), and at equilibrium, if α is the degree of ionization, the concentrations will be:
- [HX] = C(1 - α)
- [H⁺] = Cα
- [X⁻] = Cα

4. **Relate the concentration of hydrogen ions to the degree of ionization:**
- From the previous step, we have:
\[ [H^+] = Cα \]
- Substituting the known values:
\[ 10^{-5} = 0.002α \]

5. **Solve for α (degree of ionization):**
- Rearranging the equation gives:
\[ α = \frac{10^{-5}}{0.002} \]
- Calculating α:
\[ α = \frac{10^{-5}}{2 \times 10^{-3}} = 0.005 \]

6. **Calculate the percentage ionization:**
- Percentage ionization = α × 100%
- Therefore:
\[ \text{Percentage ionization} = 0.005 × 100 = 0.5\% \]

### Final Answer:
The percentage ionization of the acid in the solution is **0.5%**.

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