If the equilibrium constant of the reaction of weak acid HA with strong base is 10^9, then pH of

step 1 We're given a weak acid and a strong base in the question. For acid, we can write its dissociati

If the equilibrium constant of the reaction of weak acid HA...

Key Concepts

Acid Dissociation Constant (Ka)

Ka is the equilibrium constant for the dissociation of an acid (HA) into its conjugate base (A–) and a proton (H+). It is a measure of the strength of the acid; a higher Ka indicates a stronger acid, whereas a lower Ka signifies a weaker acid. In this context, determining Ka is essential because it governs the behavior of the acid and its conjugate base in water.

Neutralization Reaction Equilibrium

The neutralization reaction between a weak acid (HA) and a strong base (like NaOH) is represented by HA + OH– ? A– + H2O. The equilibrium constant for this reaction, when provided, allows us to relate the acid dissociation constant (Ka) of HA to the overall driving force of the reaction when it is neutralized by a strong base.

Relationship between Ka and Neutralization Constant

When a weak acid reacts with OH–, the overall reaction can be seen as the sum of the acid's dissociation (HA ? H+ + A–) and the formation of water from H+ and OH– (H+ + OH– ? H2O). This relationship yields an equilibrium constant for the neutralization reaction given by K = Ka/Kw, where Kw is the ionization constant of water. Knowing this relationship allows the determination of Ka when K and Kw are known.

Conjugate Base Hydrolysis

When the salt of a weak acid (such as NaA) is dissolved in water, its conjugate base (A–) reacts with water in a hydrolysis reaction: A– + H2O ? HA + OH–. The extent of this reaction is governed by the base hydrolysis constant (Kb), which relates to the acid’s Ka by the relation Kb = Kw/Ka. This hydrolysis is responsible for the basic pH observed in solutions of salts derived from weak acids.

pH Calculation from Salt Hydrolysis

The pH of a solution containing a salt like NaA can be determined by setting up the equilibrium for the hydrolysis of the conjugate base (A–) in water and using the relation Kb = [HA][OH–]/[A–]. By solving for the concentration of hydroxide ions generated, one can find the pOH and then convert it to pH. In the example provided, using the determined Kb leads to a calculated pH of about 9 for a 0.1 M solution.

Transcript

00:01 We're given a weak acid and a strong base in the question.

00:05 For acid, we can write its dissociation, which is incomplete, and its dissociation constant is according to this equation, as you see here in top.

00:16 And for the reaction with a base, we can write an equilibrium constant.

00:21 Okay, and the equilibrium constant, and for the base, don't forget that, we could put only the cation or anion.

00:27 However, as you see here, i've considered this relation, sodium in the form of sodium hydroxide.

00:35 So the water is produced and the sodium salt of the anion of the base.

00:42 Okay.

00:44 And what we have here is this relation that you can put also include the anion, concentration of anion instead of the salt, and either the cadine or sodium or hydroxide for the base.

00:58 Well, there's an advantage if you put hydroxide concentration here and the anion concentration, you would have common, first of all, common concentration here with the acid dissociation.

01:12 And you have hydrogen and you have hydroxide, hydroxide.

01:16 So that means you would find a relationship if you consider these two equilibrium constants for dissociation of acid.

01:30 It independently and for its equilibrium with the base.

01:35 And this is what i've done.

01:36 So i've divided these two values, considering these concentrations that i mentioned lately...

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