Prove that: Ka * Kb = Kw and pKa + pKb = 14 Using the equations below. ?? = [HA][OH-] b[A AT1 2H20 -H30+ + OH-
The Correct Answer and Explanation is :
To prove Ka⋅Kb=KwK_a \cdot K_b = K_w and pKa+pKb=14pK_a + pK_b = 14, let’s use the following information and equations:
- Definition of KaK_a and KbK_b:
- Ka=[H+][A−][HA]K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}, where [HA][\text{HA}] is the concentration of the weak acid, [A−][\text{A}^-] is its conjugate base, and [H+][\text{H}^+] is the hydronium ion concentration.
- Kb=[OH−][HB+][B]K_b = \frac{[\text{OH}^-][\text{HB}^+]}{[\text{B}]}, where [B][\text{B}] is the concentration of the weak base, [HB+][\text{HB}^+] is its conjugate acid, and [OH−][\text{OH}^-] is the hydroxide ion concentration.
- Water autoionization constant:
- Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-] at 25°C, and pKw=−logKw=14\text{p}K_w = -\log K_w = 14.
Derivation of Ka⋅Kb=KwK_a \cdot K_b = K_w
- For a weak acid HA\text{HA}, the conjugate base A−\text{A}^- reacts with water: A−+H2O↔HA+OH−\text{A}^- + \text{H}_2\text{O} \leftrightarrow \text{HA} + \text{OH}^- Here, Kb=[OH−][HA][A−]K_b = \frac{[\text{OH}^-][\text{HA}]}{[\text{A}^-]}.
- Multiplying the expressions for KaK_a and KbK_b: Ka⋅Kb=([H+][A−][HA])⋅([OH−][HA][A−])K_a \cdot K_b = \left( \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \right) \cdot \left( \frac{[\text{OH}^-][\text{HA}]}{[\text{A}^-]} \right) Simplify: Ka⋅Kb=[H+][OH−]=KwK_a \cdot K_b = [\text{H}^+][\text{OH}^-] = K_w
Derivation of pKa+pKb=14pK_a + pK_b = 14
- By definition, pKa=−logKapK_a = -\log K_a and pKb=−logKbpK_b = -\log K_b.
- Take the negative logarithm of both sides of Ka⋅Kb=KwK_a \cdot K_b = K_w: −log(Ka⋅Kb)=−logKw-\log(K_a \cdot K_b) = -\log K_w Using log properties: −logKa−logKb=−logKw-\log K_a – \log K_b = -\log K_w Substituting pKw=14\text{p}K_w = 14: pKa+pKb=14pK_a + pK_b = 14
Explanation (300 Words)
The relationships Ka⋅Kb=KwK_a \cdot K_b = K_w and pKa+pKb=14pK_a + pK_b = 14 stem from the equilibrium principles of weak acids and bases and water’s autoionization constant. A weak acid (HA\text{HA}) dissociates in water to form [H+][\text{H}^+] and its conjugate base (A−\text{A}^-), described by KaK_a. Similarly, the conjugate base (A−\text{A}^-) reacts with water to produce [OH−[\text{OH}^- and HA\text{HA}, described by KbK_b.
The product Ka⋅KbK_a \cdot K_b links the dissociation of an acid and its conjugate base through the relationship Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-], a constant value at a given temperature (25°C, Kw=1×10−14K_w = 1 \times 10^{-14}). The logarithmic expressions, pKa=−logKapK_a = -\log K_a and pKb=−logKbpK_b = -\log K_b, reflect the inverse relationship between an acid’s strength and its conjugate base’s strength. Their sum always equals 1414, corresponding to the water ionization constant (pKw=14\text{p}K_w = 14).
This mathematical proof underpins acid-base equilibrium concepts, demonstrating how KaK_a, KbK_b, and KwK_w are interrelated. It emphasizes that strong acids have weak conjugate bases and vice versa, maintaining balance in aqueous systems.