The pKb of water is _______. (a) 1 (b) 7 (c) 14 (d) not defined (e) none of the above

Verified step by step guidance

Step 1: Understand the concept of pKb. The pKb is the negative logarithm of the base dissociation constant (Kb) of a substance. It is used to express the strength of a base in solution.

Step 2: Recall the autoionization of water. Water can act as both an acid and a base, and it undergoes autoionization to form hydronium (H3O+) and hydroxide (OH-) ions.

Step 3: Consider the equilibrium constant for water's autoionization. The equilibrium constant for this reaction is known as the ion product of water (Kw), which is equal to 1.0 x 10^-14 at 25°C.

Step 4: Relate pKw to pKa and pKb. For water, pKw = pKa + pKb. Since pKw is 14 at 25°C and water is neutral, pKa and pKb are equal, each being 7.

Step 5: Conclude that the pKb of water is 7, based on the relationship between pKw, pKa, and pKb.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

pKb and pKa

pKb is a measure of the basicity of a substance, specifically the negative logarithm of the base dissociation constant (Kb). It indicates how readily a base accepts protons in solution. The relationship between pKa and pKb is crucial, as they are related through the equation pKa + pKb = 14 at 25°C, which is essential for understanding acid-base equilibria.

Water as an Amphoteric Substance

Water (H2O) is considered an amphoteric substance because it can act both as an acid and a base. This dual behavior allows water to participate in various acid-base reactions, contributing to its role in chemical equilibria. Understanding this property is vital for grasping the context of pKb values in relation to water.

Dissociation of Water

The dissociation of water into hydrogen ions (H+) and hydroxide ions (OH-) is a fundamental concept in chemistry, represented by the equilibrium expression Kw = [H+][OH-]. At 25°C, Kw is 1.0 x 10^-14, leading to the conclusion that the pH and pOH of pure water are both 7. This equilibrium is essential for understanding the pKb of water and its implications in acid-base chemistry.

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Related Practice

Textbook Question

The amino acid glycine (H₂N–CH₂–COOH) can participate in the following equilibria in water:

H₂N–CH₂–COOH + H₂O ⇌ H₂N–CH₂–COO^– + H₃O⁺ K_a = 4.3 × 10^-3

H₂N–CH₂–COOH + H₂O⇌ ⁺H₃N–CH₂–COOH + OH^- K_b = 6.0 × 10^-5

(a) Use the values of Ka and K_b to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: H2N–CH2–COOH ⇌ ⁺H₃N–CH₂–COO^–

Textbook Question

The amino acid glycine (H₂N–CH₂–COOH) can participate in the following equilibria in water:

H₂N–CH₂–COOH + H₂O ⇌ H₂N–CH₂–COO^– + H₃O⁺ K_a = 4.3 × 10^-3

H₂N–CH₂–COOH + H₂O⇌ ⁺H₃N–CH₂–COOH + OH^- K_b = 6.0 × 10^-5

(b) What is the pH of a 0.050 M aqueous solution of glycine?

Textbook Question

The volume of an adult's stomach ranges from about 50 mL when empty to 1 L when full. If the stomach volume is 400 mL and its contents have a pH of 2, how many moles of H+ does the stomach contain? Assuming that all the H+ comes from HCl, how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?