Answer

Step 1: Calculate Equilibrium [H⁺]

[H⁺] = 10⁻².⁶ = 0.00251 mol dm⁻³

Step 2: ICE Table

	HA	⇌	H⁺	+	A⁻
Initial	0.5		0		0
Change	-x		+x		+x
Equilibrium	0.5 – 0.00251		0.00251		0.00251

x = [H⁺] = 0.00251

Step 3: Calculate Kₐ
[HA]_eq = 0.5 – 0.00251 = 0.4975 mol dm⁻³
Kₐ = [H⁺][A⁻] / [HA]_eq
Kₐ = (0.00251)² / 0.4975
Kₐ = 1.27 × 10⁻⁵ mol dm⁻³

Answer

Step 1: Calculate Equilibrium [H⁺]

[H⁺] = 10⁻¹.⁰⁸ = 0.0832 mol dm⁻³

Step 2: ICE Table

	HA	⇌	H⁺	+	A⁻
Initial	0.5		0		0
Change	-x		+x		+x
Equilibrium	0.5 – 0.0832		0.0832		0.0832

x = [H⁺] = 0.0832

Step 3: Calculate Kₐ
[HA]_eq = 0.5 – 0.0832 = 0.4168 mol dm⁻³
Kₐ = (0.0832)² / 0.4168
Kₐ = 1.66 × 10⁻² mol dm⁻³

Answer

Step 1: Calculate Equilibrium [H⁺]

[H⁺] = 10⁻¹.⁶⁵ = 0.0224 mol dm⁻³

Step 2: ICE Table

	HA	⇌	H⁺	+	A⁻
Initial	0.1		0		0
Change	-x		+x		+x
Equilibrium	0.1 – 0.0224		0.0224		0.0224

x = [H⁺] = 0.0224

Step 3: Calculate Kₐ
[HA]_eq = 0.1 – 0.0224 = 0.0776 mol dm⁻³
Kₐ = (0.0224)² / 0.0776
Kₐ = 6.47 × 10⁻³ mol dm⁻³

Answer

Part 1: Find Kₐ

[H⁺] = 10⁻³.⁵⁰ = 3.16×10⁻⁴

	HA	⇌	H⁺	+	A⁻
Initial	0.02		0		0
Change	-x		+x		+x
Equilibrium	0.0197		x		x

x = 3.16×10⁻⁴. [HA]_eq = 0.02 – 3.16×10⁻⁴ = 0.0197

Kₐ = (3.16×10⁻⁴)² / 0.0197 = 5.07×10⁻⁶

Part 2: Calculate new pH at 0.8 mol dm⁻³

Since Kₐ is very small, we can assume [HA]_eq ≈ 0.8.
[H⁺] = √(Kₐ × 0.8) = √(5.07×10⁻⁶ × 0.8)
[H⁺] = 2.01×10⁻³ mol dm⁻³
pH = -log(2.01×10⁻³) = 2.70

Answer

Part 1: Find Kₐ

[H⁺] = 10⁻¹.⁸⁰ = 0.0158

	HA	⇌	H⁺	+	A⁻
Initial	0.100		0		0
Equilibrium	0.0842		0.0158		0.0158

[HA]_eq = 0.1 – 0.0158 = 0.0842

Kₐ = (0.0158)² / 0.0842 = 2.97×10⁻³

Part 2: Calculate Concentration for pH 1.00

Target [H⁺] = 10⁻¹ = 0.1 mol dm⁻³
We know Kₐ = [H⁺]² / [HA]_eq
[HA]_eq = [H⁺]² / Kₐ = (0.1)² / 2.97×10⁻³ = 3.37 mol dm⁻³

Find Initial Concentration:
[HA]_initial = [HA]_eq + [H⁺] (dissociated)
[HA]_initial = 3.37 + 0.1 = 3.47 mol dm⁻³