An acidic indicator ionises as `I_(n) H hArr "In"^(-) + H^(+)` . The molecular and ions of the indicator show different colours. The indicator changed it colour if its acidic of basic form completely predominant.
Dissociation constant of an acidic indicator is `10^(-5)` . At what `p^(H)` 80% of the indicator exists in molecular form ?

Options

A. 4.4
B. 6.6
C. 5.5
D. 3.3

Answer

To find the pH at which 80% of the indicator exists in its molecular form, we can follow these steps:

### Step 1: Understand the Dissociation of the Indicator
The dissociation of the acidic indicator can be represented as:
\[ \text{HIn} \rightleftharpoons \text{In}^- + \text{H}^+ \]

### Step 2: Define the Dissociation Constant (Ka)
The dissociation constant \( K_a \) for the indicator is given as:
\[ K_a = 10^{-5} \]

### Step 3: Calculate pKa
To find \( pK_a \), we can use the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(10^{-5}) = 5 \]

### Step 4: Determine the Ratio of Ionized to Unionized Forms
We know that 80% of the indicator exists in the molecular form (HIn) and 20% in the ionized form (In^-). This gives us the ratio:
\[ \frac{[\text{In}^-]}{[\text{HIn}]} = \frac{20}{80} = \frac{1}{4} \]

### Step 5: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates the pH, pKa, and the ratio of the concentrations of the ionized and unionized forms:
\[ pH = pK_a + \log\left(\frac{[\text{In}^-]}{[\text{HIn}]}\right) \]
Substituting the values we have:
\[ pH = 5 + \log\left(\frac{1}{4}\right) \]

### Step 6: Calculate the Logarithm
We know that:
\[ \log\left(\frac{1}{4}\right) = \log(1) - \log(4) = 0 - 0.602 = -0.602 \]
Thus:
\[ pH = 5 - 0.602 = 4.398 \]

### Step 7: Round the pH Value
Rounding the pH value gives:
\[ pH \approx 4.4 \]

### Final Answer
The pH at which 80% of the indicator exists in molecular form is approximately **4.4**.

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