Question

If the dissociation constant of a base is 2 × 10–3, then the dissociation constant of its conjugate acid at 25°C will be​

Answer 1

Answer:

Ka=5*10^-12

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Answer 2

Given :

Dissociation constant of base (Kb) = 2×10⁻³

To find :

Dissociation constant of its conjugate acid

Solution :

• We know that Ka × Kb = Kw

• Ka × Kb = 10⁻¹⁴              [Kw = 10⁻¹⁴]

• Ka × 2×10⁻³ = 10⁻¹⁴  

• Ka = 10⁻¹⁴ / (2×10⁻³)

• Ka = 5×10⁻¹²

The value of dissociation constant of conjugate acid is 5×10⁻¹²