Q. weak acid + weak base

Step 1: Identify the components and the reaction

Let us define the weak acid as \( \text{HA} \) and the weak base as \( \text{B} \). The neutralization reaction between them forms a salt consisting of the conjugate acid \( \text{BH}^{+} \) and the conjugate base \( \text{A}^{-} \). The chemical equation can be written as follows.

\[ \text{HA} + \text{B} = \text{BH}^{+} + \text{A}^{-} \]

Step 2: Understand the hydrolysis of the resulting ions

Unlike strong acid or strong base neutralizations, both ions of the resulting salt undergo hydrolysis when dissolved in water. This means they react with water to produce hydronium ions \( \text{H}_3\text{O}^{+} \) and hydroxide ions \( \text{OH}^{-} \).

The conjugate acid \( \text{BH}^{+} \) reacts with water to produce hydronium.

\[ \text{BH}^{+} + \text{H}_2\text{O} = \text{B} + \text{H}_3\text{O}^{+} \]

The conjugate base \( \text{A}^{-} \) reacts with water to produce hydroxide.

\[ \text{A}^{-} + \text{H}_2\text{O} = \text{HA} + \text{OH}^{-} \]

Step 3: Define the equilibrium constants

The strength of the weak acid is given by its acid dissociation constant, \( K_a \). The strength of the weak base is given by its base dissociation constant, \( K_b \). The autoionization constant of water is \( K_w \).

Step 4: Derive the concentration of hydronium ions

Through charge balance and mass balance equations for this specific system, we can derive the hydrogen ion concentration. Assuming that the degree of hydrolysis is small, the concentration of the hydronium ion \( \left[ \text{H}^{+} \right] \) can be approximated by the following formula.

\[ \left[ \text{H}^{+} \right] = \sqrt{ \frac{ K_w \cdot K_a }{ K_b } } \]

Step 5: Convert the concentration to the pH formula

To find the pH, we take the negative base-ten logarithm of both sides of the equation. Remember that \( \text{pH} = -\log_{10}\left( \left[ \text{H}^{+} \right] \right) \) and \( pK = -\log_{10}(K) \).

\[ \begin{aligned} \text{pH} &= -\log_{10}\left( \left( \frac{ K_w \cdot K_a }{ K_b } \right)^{\frac{1}{2}} \right) \\ &= \frac{1}{2} \left( -\log_{10}(K_w) – \log_{10}(K_a) + \log_{10}(K_b) \right) \\ &= \frac{1}{2} \left( pK_w + pK_a – pK_b \right) \end{aligned} \]

At standard room temperature, \( pK_w \) is exactly equal to 14. Substituting this value gives the final working formula.

\[ \text{pH} = 7 + \frac{1}{2} \left( pK_a – pK_b \right) \]

Step 6: Analyze the final solution

The final pH of the solution depends entirely on the relative values of \( K_a \) and \( K_b \).

If \( K_a > K_b \), the acid is stronger than the base. The term \( pK_a – pK_b \) becomes negative, making the \( \text{pH} < 7 \). The solution is acidic.

If \( K_a < K_b \), the base is stronger than the acid. The term \( pK_a – pK_b \) becomes positive, making the \( \text{pH} > 7 \). The solution is basic.

If \( K_a = K_b \), the acid and base are of equal strength. The term \( pK_a – pK_b \) becomes zero, making the \( \text{pH} = 7 \). The solution is perfectly neutral.