Question

A weak acid HA has degree of dissociation x. Which option gives the correct expression of \(pH - pK_a\)?

• A. log\(\left(\frac{1-x}{x}\right)\)
• B. 0
• C. log\(\left(\frac{x}{1-x}\right)\)
• D. log(1 + 2x)

Hint:

For weak acids, dissociation can be quantified by the degree of dissociation x, which influences the pH relative to its \(pK_a\).

Answer 1

The Correct Option is A

We begin with the more accurate expression for the dissociation of a weak acid:

$$ K_a = \frac{Cx^2}{1 - x} $$ where $ C $ is the initial concentration of the acid, and $ x $ is the degree of dissociation.

The hydrogen ion concentration is given by:

$$ [H^+] = Cx $$ So, $$ pH = -\log(Cx) $$

Similarly, the $ pK_a $ is defined as:

$$ pK_a = -\log(K_a) = -\log\left(\frac{Cx^2}{1 - x}\right) $$

Now, we calculate $ pH - pK_a $:

$$ pH - pK_a = -\log(Cx) + \log\left(\frac{Cx^2}{1 - x}\right) = \log\left(\frac{Cx^2}{Cx(1 - x)}\right) = \log\left(\frac{x}{1 - x}\right) $$

This gives us the general relation:

$$ pH - pK_a = \log\left(\frac{x}{1 - x}\right) $$

If $ pH = pK_a $, then:

$$ \log\left(\frac{x}{1 - x}\right) = 0 \Rightarrow \frac{x}{1 - x} = 1 \Rightarrow x = 1 - x \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2} $$

Thus, when the degree of dissociation $ x = 0.5 $, the pH equals the $ pK_a $, which corresponds to the midpoint of the titration curve.

Final Answer:
The correct expression for $ pH - pK_a $ without approximation is:

$$ \ \log\left(\frac{x}{1 - x}\right) $$

Answer 2

Step 1: Understand the question.
We are dealing with a weak acid HA having a degree of dissociation \( x \). We need to derive the relationship between \( pH \) and \( pK_a \) in terms of \( x \).

Step 2: Expression for dissociation of a weak acid.
For a weak acid \( HA \), the equilibrium reaction is:
\[ HA \rightleftharpoons H^+ + A^- \] If the initial concentration of the acid is \( C \), then at equilibrium:
\[ [H^+] = Cx, \quad [A^-] = Cx, \quad [HA] = C(1 - x) \]

Step 3: Write the expression for the acid dissociation constant.
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(Cx)(Cx)}{C(1 - x)} = \frac{Cx^2}{1 - x} \]

Step 4: Take negative logarithm of both sides.
\[ pK_a = -\log K_a = -\log \left( \frac{Cx^2}{1 - x} \right) = -\log C - 2\log x + \log(1 - x) \]

Since \( pH = -\log [H^+] = -\log (Cx) = -\log C - \log x \),
Subtract \( pK_a \) from \( pH \):
\[ pH - pK_a = (-\log C - \log x) - (-\log C - 2\log x + \log(1 - x)) \] Simplifying:
\[ pH - pK_a = \log(1 - x) - \log x = \log\left( \frac{1 - x}{x} \right) \]

Step 5: Conclusion.
The correct expression relating \( pH \) and \( pK_a \) for a weak acid with degree of dissociation \( x \) is:
\[ pH - pK_a = \log\left( \frac{1 - x}{x} \right) \]

Final Answer:
log\(\left(\frac{1-x}{x}\right)\)