뒤로Ka and Kb quiz #1
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1/28뒤로Skip to main content뒤로How is the base ionization constant (Kb) defined in terms of equilibrium concentrations?
Kb quantifies a base's ability to accept protons and is defined by the equilibrium expression: Kb = [BH+][OH−]/[B], where B is the base. Which statement correctly describes the Ka of a weak acid?
The Ka of a weak acid is less than 1, indicating that the acid only partially ionizes in water. The Kb of NH3 is 1.8 × 10⁻⁵. What is the Ka of NH4⁺?
Ka = Kw / Kb = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. What is the relationship between Ka and Kb values? How can you determine Ka knowing Kb?
Ka × Kb = Kw (1.0 × 10⁻¹⁴ at 25°C). To find Ka, use Ka = Kw / Kb. The Ka of HCN is 6.2 × 10⁻¹⁰. What is the Kb value for CN⁻ at 25°C?
Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (6.2 × 10⁻¹⁰) = 1.6 × 10⁻⁵. What is Kb for the conjugate base of HCN (Ka = 4.9 × 10⁻¹⁰)?
Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (4.9 × 10⁻¹⁰) = 2.0 × 10⁻⁵. Which of these salts will be acidic? (Use the Ka/Kb table from general chemistry)
Salts formed from a strong acid and a weak base will be acidic, as the conjugate acid of the weak base hydrolyzes to produce H⁺. Given that Ka for HF is 6.3 × 10⁻⁴ at 25°C, what is the Kb for F⁻?
Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (6.3 × 10⁻⁴) = 1.6 × 10⁻¹¹. If the Ka of a monoprotic weak acid is 5.9 × 10⁻⁶, what does this indicate about the acid?
The acid is weak and only partially ionizes in water, as its Ka is much less than 1. A 0.200 M solution of a weak acid has a pH of 3.15. What is the value of Ka for the acid?
First, [H₃O⁺] = 10⁻³.¹⁵ = 7.1 × 10⁻⁴ M. Assume x ≈ [H₃O⁺], Ka = x² / (0.200 - x) ≈ (7.1 × 10⁻⁴)² / 0.200 = 2.5 × 10⁻⁶. If you multiply the Ka for HF by the Kb value for F⁻, what do you get?
You get Kw, the ionization constant for water (1.0 × 10⁻¹⁴ at 25°C). What is Ka for the conjugate acid of NH₂⁻ (Kb = 1.8 × 10⁻⁵)?
Ka = Kw / Kb = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. HA is a weak acid. Which equilibrium corresponds to the equilibrium constant Kb for A⁻?
A⁻ + H₂O ⇌ HA + OH⁻; Kb = [HA][OH⁻]/[A⁻]. If the Ka of a monoprotic weak acid is 3.1 × 10⁻⁶, what does this tell you?
The acid is weak and only partially dissociates in water. What is Ka for the conjugate acid of C₅H₅N (Kb = 1.7 × 10⁻⁹)?
Ka = Kw / Kb = (1.0 × 10⁻¹⁴) / (1.7 × 10⁻⁹) = 5.9 × 10⁻⁶. The Ka for HCN is 4.9 × 10⁻¹⁰. What is the value of Kb for CN⁻?
Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (4.9 × 10⁻¹⁰) = 2.0 × 10⁻⁵. A 0.145 M solution of a weak acid has a pH of 2.75. What is the value of Ka for the acid?
[H₃O⁺] = 10⁻².⁷⁵ = 1.8 × 10⁻³ M. Ka ≈ (1.8 × 10⁻³)² / 0.145 = 2.2 × 10⁻⁵. The pKa of HF is 3.2. Determine the pKb of its conjugate base.
pKa + pKb = 14; pKb = 14 - 3.2 = 10.8. What is the Ka of an acid if a 0.500 M solution contains 1.70 × 10⁻⁴ M H₃O⁺?
Ka = (1.70 × 10⁻⁴)² / (0.500 - 1.70 × 10⁻⁴) ≈ 5.8 × 10⁻⁸. If the Kb of a weak base is 8.9 × 10⁻⁶, what is the Ka of its conjugate acid?
Ka = Kw / Kb = (1.0 × 10⁻¹⁴) / (8.9 × 10⁻⁶) = 1.1 × 10⁻⁹. What is Ka for the conjugate acid of CH₃NH₂ (Kb = 4.4 × 10⁻⁴)?
Ka = Kw / Kb = (1.0 × 10⁻¹⁴) / (4.4 × 10⁻⁴) = 2.3 × 10⁻¹¹. If the pH at the half-titration point of a monoprotic weak acid is 4.2, what is the Ka of the acid?
At half-titration, pH = pKa; Ka = 10⁻⁴.² = 6.3 × 10⁻⁵. What is the Ka reaction of HCN?
HCN + H₂O ⇌ CN⁻ + H₃O⁺; Ka = [CN⁻][H₃O⁺]/[HCN]. Determine the dissociation constants for the acids.
The dissociation constant (Ka) for an acid is calculated using the equilibrium concentrations: Ka = [A⁻][H₃O⁺]/[HA]. Write the Ka reaction of HCN.
HCN + H₂O ⇌ CN⁻ + H₃O⁺; Ka = [CN⁻][H₃O⁺]/[HCN]. The Ka of propanoic acid is 1.34 × 10⁻⁵. What does this value indicate?
Propanoic acid is a weak acid, as its Ka is much less than 1. What is the Ka of HF?
The Ka of HF is 6.3 × 10⁻⁴. What is the Ka of HCN?
The Ka of HCN is 4.9 × 10⁻¹⁰.
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