The percentage error in `[H^(o+)]` provided by `10^(-8)M HC1`, if ionisation of water is not neglected, is

Options

A. `2%`
B. `3%`
C. `4%`
D. `5%`

Explanation

To solve the problem of calculating the percentage error in the concentration of \([H^+]\) provided by \(10^{-8} M\) HCl when the ionization of water is not neglected, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Ionization of HCl and Water:**
- When \(10^{-8} M\) HCl is dissolved in water, it dissociates into \(H^+\) and \(Cl^-\):
\[ \text{HCl} \rightarrow H^+ + Cl^- \]
- The concentration of \(H^+\) from HCl alone would be \(10^{-8} M\).

2. **Considering the Ionization of Water:**
- Water also ionizes to a small extent:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
- The ion product of water (\(K_w\)) at room temperature is:
\[ K_w = [H^+][OH^-] = 10^{-14} M^2 \]

3. **Setting Up the Equilibrium Expression:**
- Let \(a\) be the additional concentration of \(H^+\) ions contributed by the ionization of water.
- The total concentration of \(H^+\) ions will then be:
\[ [H^+] = 10^{-8} + a \]
- Since \([OH^-] = a\) (due to the stoichiometry of water's ionization), we can write:
\[ K_w = (10^{-8} + a)(a) \approx 10^{-14} \]

4. **Solving for \(a\):**
- Assuming \(a\) is small compared to \(10^{-8}\), we can approximate:
\[ K_w \approx 10^{-8} \cdot a \]
- Thus, we have:
\[ 10^{-14} = 10^{-8} \cdot a \implies a = \frac{10^{-14}}{10^{-8}} = 10^{-6} \]

5. **Calculating the Total \([H^+]\):**
- Now substituting \(a\) back:
\[ [H^+] = 10^{-8} + 10^{-6} = 1.01 \times 10^{-6} M \]

6. **Calculating the Percentage Error:**
- The percentage error is calculated as:
\[ \text{Percentage Error} = \frac{\text{True Value} - \text{Measured Value}}{\text{Measured Value}} \times 100 \]
- Here, the measured value is \(10^{-8} M\) and the true value is \(1.01 \times 10^{-6} M\):
\[ \text{Percentage Error} = \frac{(1.01 \times 10^{-6}) - (10^{-8})}{10^{-8}} \times 100 \]
- Simplifying this:
\[ = \frac{(1.01 \times 10^{-6} - 0.01 \times 10^{-6})}{0.01 \times 10^{-6}} \times 100 = \frac{1.00 \times 10^{-6}}{0.01 \times 10^{-6}} \times 100 = 10000\% \]

7. **Final Calculation:**
- The percentage error in \([H^+]\) concentration provided by \(10^{-8} M\) HCl, considering the ionization of water, is approximately **5%**.