The ionisation constant of an acid base indicator (a weak acid) is `1.0 xx 10^(-6)`. The ionised form of the indicator is red and unionised form is blue. The p`H` change required to alter the colour of indicator from `80%` red to 20% red is

Options

A. `0.80`
B. `1.20`
C. `1.40`
D. `2.00`

Explanation

To solve the problem, we need to determine the pH change required to alter the color of the indicator from 80% red (ionized form) to 20% red. The ionization constant (Ka) of the indicator is given as \(1.0 \times 10^{-6}\).

### Step-by-step Solution:

1. **Identify the Ionization Constant**:
The ionization constant (Ka) of the indicator is given as:
\[ K_a = 1.0 \times 10^{-6} \]
We can find the pKa using the formula:
\[ pK_a = -\log(K_a) = -\log(1.0 \times 10^{-6}) = 6 \]

2. **Initial Condition**:
Initially, the indicator is 80% ionized (red) and 20% unionized (blue). We can denote:
- Concentration of ionized form (I⁻) = 80
- Concentration of unionized form (HIn) = 20
The total concentration is 100, so the ratio of the ionized form to the unionized form is:
\[ \frac{[I^-]}{[HIn]} = \frac{80}{20} = 4 \]

3. **Calculate Initial pH**:
Using the Henderson-Hasselbalch equation:
\[ pH_{initial} = pK_a + \log\left(\frac{[I^-]}{[HIn]}\right) \]
Substituting the values:
\[ pH_{initial} = 6 + \log(4) \]
Since \(\log(4) = 2 \log(2)\):
\[ pH_{initial} = 6 + 2 \log(2) \]

4. **Final Condition**:
In the final condition, the indicator is 20% ionized (red) and 80% unionized (blue):
- Concentration of ionized form (I⁻) = 20
- Concentration of unionized form (HIn) = 80
The ratio now is:
\[ \frac{[I^-]}{[HIn]} = \frac{20}{80} = \frac{1}{4} \]

5. **Calculate Final pH**:
Again using the Henderson-Hasselbalch equation:
\[ pH_{final} = pK_a + \log\left(\frac{[I^-]}{[HIn]}\right) \]
Substituting the values:
\[ pH_{final} = 6 + \log\left(\frac{1}{4}\right) \]
Since \(\log\left(\frac{1}{4}\right) = -\log(4) = -2 \log(2)\):
\[ pH_{final} = 6 - 2 \log(2) \]

6. **Calculate the Change in pH**:
The change in pH (\(\Delta pH\)) is given by:
\[ \Delta pH = pH_{initial} - pH_{final} \]
Substituting the values:
\[ \Delta pH = \left(6 + 2 \log(2)\right) - \left(6 - 2 \log(2)\right) \]
Simplifying:
\[ \Delta pH = 4 \log(2) \]

7. **Calculate the Numerical Value**:
Using \(\log(2) \approx 0.301\):
\[ \Delta pH \approx 4 \times 0.301 = 1.204 \]

### Final Answer:
The pH change required to alter the color of the indicator from 80% red to 20% red is approximately **1.2**.