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IIydrolysis constant of a salt of weak acid and weak basc is inversely proportional to (1) Dissociation constant of weak acid (2) Dissociation constant of weak base (3) lonic product of water (4) Dissociation constant of both weak acid and weak base

Short Answer

Expert verified
4) Dissociation constant of both weak acid and weak base

Step by step solution

01

Understand the Concept

The hydrolysis constant of a salt formed from a weak acid and a weak base is the equilibrium constant for the hydrolysis reaction of the salt in water.
02

Write the Reaction

Consider a salt formed from a weak acid (HA) and a weak base (BOH). The salt dissociates completely in water to give A^- and B^+. These ions can hydrolyze in water: A^- + H2O ↔ HA + OH^- and B^+ + H2O ↔ BOH + H^+.
03

Define Hydrolysis Constant

The hydrolysis constant (K_h) can be defined in terms of the equilibrium constants for the weak acid (K_a) and the weak base (K_b) and the ionic product of water (K_w): K_h = K_w / (K_a * K_b).
04

Analyze the Proportionality

From the formula K_h = K_w / (K_a * K_b), it is clear that the hydrolysis constant is inversely proportional to the dissociation constants of both the weak acid (K_a) and the weak base (K_b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weak Acid
A weak acid only partially dissociates in water. This means that not all acid molecules break apart into their ions when dissolved.
For example, acetic acid (CH3COOH) is a weak acid. When dissolved in water, it does not completely dissociate into CH3COO^- and H^+ ions.
Instead, an equilibrium is established, meaning that both the undissociated acid and its ions coexist in the solution.
The extent of this dissociation is quantified by the acid dissociation constant, denoted as K_a.
Mathematically: \( K_a = \frac{[H^+][A^-]}{[HA]} \)
This equation tells us how much of the acid is dissociated at equilibrium.
The smaller the value of K_a, the weaker the acid because fewer molecules dissociate.
Weak Base
A weak base is similar to a weak acid in that it does not fully dissociate in water.
When dissolved, only some of the base molecules accept protons (H^+) from water to yield hydroxide ions (OH^-) and the conjugate acid.
For instance, ammonia (NH3) in water partially dissociates to form ammonium ions (NH4^+) and hydroxide ions (OH^-).
This partial dissociation is characterized by the base dissociation constant, K_b.
The formula to represent this equilibrium is given by: \( K_b = \frac{[BH^+][OH^-]}{[B]} \)
The lower the value of K_b, the weaker the base, indicating that fewer base molecules are accepting protons.
Ionic Product of Water
The ionic product of water (K_w) is a special equilibrium constant that applies to water.
It represents the product of the concentrations of hydrogen ions (H^+) and hydroxide ions (OH^-) in water at a particular temperature.
At 25 degrees Celsius, K_w is: \( K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \)
This value reflects the fact that water constantly undergoes slight self-ionization to produce these ions.
K_w is essential for calculating the hydrolysis constant of salts derived from weak acids and bases, as it factors into the relationship.
Dissociation Constant
Dissociation constants provide a quantitative measure of the strength of acids and bases.
For acids, the dissociation constant K_a describes the ratio of dissociated ions to undissociated molecules at equilibrium.
Similarly, for bases, K_b represents the ratio indicating base ionization.
These constants are critical to understanding hydrolysis reactions because they describe how readily acids and bases dissociate in water.
In summary, for a salt formed from a weak acid and weak base, the hydrolysis constant (K_h) is related to the ionic product of water (K_w) and the dissociation constants (K_a and K_b) via the formula: \( K_h = \frac{K_w}{K_a \cdot K_b} \).
This equation shows that K_h is inversely proportional to both K_a and K_b, illustrating how changes in the dissociation constants impact the hydrolysis constant.

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Most popular questions from this chapter

\begin{aligned} &\text { The hydrolysis constant of a salt of weak acid }\\\ &\left(K_{\mathrm{a}}=2 \times 10^{-6}\right) \text { and of a weak basc }\left(K_{\mathrm{b}}=5 \times 10^{-7}\right) \text { is } \end{aligned}(1) 10 (2) \(10^{2}\) (3) \(2 \times 10^{2}\) (4) \(5 \times 10^{3}\)

\(\mathrm{K}_{\mathrm{c}}\) for \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\) is 10 at \(25^{\circ} \mathrm{C}\). If a con- tainer contains \(1,2,3,4\) moles per litre of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and D, respectively at \(25^{\circ} \mathbf{C}\) the reaction shall (1) Procecd from left to right (2) Procecd from right to left (3) Be at cquilibrium (4) Nonc

For which reaction high pressure and high temperature is helpful in obtaining a high cquilibrium yicld? (1) \(2 \mathrm{NF}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \quad 54.40 \mathrm{kcal}\) (2) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{II}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NI}_{3}(\mathrm{~g})+22.08 \mathrm{kcal}\) (3) \(\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClO}_{3}(\mathrm{~g}) \quad 49.4 \mathrm{kcal}\) (4) \(2 \mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Cl}_{2}(\mathrm{~g}) \times 7 \mathrm{O}_{2}(\mathrm{~g})+126.8 \mathrm{kcal}\)

Hydrochloric acid is a stronger acid than acetic acid because (1) It can neutralize large quantity of alkali. (2) It can corrode anything it comes in contact. (3) It ionizes completely into ions in an aqueous solution. (4) It ionizes partially into ions in aqueous solution.

Which cquilibrium can be described as Lcwis acid basc reaction but not Bronsted acid basc reaction? (1) \(\mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\) (2) \(2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{NH}_{4}^{+}+\mathrm{SO}_{4}^{2-}\) (3) \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NH}_{3} \rightarrow \mathrm{CH}_{3} \mathrm{COONH}_{4}\) (4) \(\left[\mathrm{Cu}\left(\mathrm{II}_{2} \mathrm{O}\right)_{4}\right]^{21}+4 \mathrm{NII}_{3} \rightarrow\left[\mathrm{Cu}\left(\mathrm{NII}_{3}\right)_{4}\right]^{2}+4 \mathrm{II}_{2} \mathrm{O}\)

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