What fraction of an indicator `H"in"` is in the basic form at a `pH` of `6` if `pK_(a)` of the indicator is `5`?

Options

A. `1/2`
B. `1/11`
C. `10/11`
D. `1/10`

Answer

To find the fraction of the indicator \( H^{in} \) that is in the basic form at a pH of 6, given that the \( pK_a \) of the indicator is 5, we can use the Henderson-Hasselbalch equation. Here’s a step-by-step solution:

### Step 1: Understand the relationship between \( pH \), \( pK_a \), and the concentrations of the forms of the indicator.
The Henderson-Hasselbalch equation is given by:

\[ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \]

where:
- \( [A^-] \) is the concentration of the basic form (ionized form, \( H^{in-} \))
- \( [HA] \) is the concentration of the acidic form (non-ionized form, \( H^{in} \))

### Step 2: Substitute the known values into the equation.
Given:
- \( pH = 6 \)
- \( pK_a = 5 \)

Substituting these values into the equation:

\[ 6 = 5 + \log \left( \frac{[H^{in-}]}{[H^{in}]} \right) \]

### Step 3: Solve for the ratio of the concentrations.
Rearranging the equation gives:

\[ 6 - 5 = \log \left( \frac{[H^{in-}]}{[H^{in}]} \right) \]

This simplifies to:

\[ 1 = \log \left( \frac{[H^{in-}]}{[H^{in}]} \right) \]

To eliminate the logarithm, we exponentiate both sides:

\[ 10^1 = \frac{[H^{in-}]}{[H^{in}]} \]

This means:

\[ \frac{[H^{in-}]}{[H^{in}]} = 10 \]

### Step 4: Determine the fraction of the indicator in the basic form.
Let \( [H^{in}] = x \) and \( [H^{in-}] = 10x \). The total concentration of the indicator is:

\[ [H^{in}] + [H^{in-}] = x + 10x = 11x \]

The fraction of the indicator in the basic form is:

\[ \text{Fraction in basic form} = \frac{[H^{in-}]}{[H^{in}] + [H^{in-}]} = \frac{10x}{11x} = \frac{10}{11} \]

### Final Answer:
The fraction of the indicator \( H^{in} \) that is in the basic form at a pH of 6 is \( \frac{10}{11} \).

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