What is the Kb of ammonia (NH3) if the pKa of its conjugate acid (NH4+) is 9.25?

Practice Questions

Q1
What is the Kb of ammonia (NH3) if the pKa of its conjugate acid (NH4+) is 9.25?
  1. 1.8 x 10^-5
  2. 5.6 x 10^-10
  3. 1.0 x 10^-14
  4. 3.2 x 10^-5

Questions & Step-by-Step Solutions

What is the Kb of ammonia (NH3) if the pKa of its conjugate acid (NH4+) is 9.25?
  • Step 1: Understand that Kb is the base dissociation constant for ammonia (NH3).
  • Step 2: Know that the relationship between Kb and Ka (the acid dissociation constant) is given by the formula: Kb = Kw / Ka.
  • Step 3: Identify Kw, which is the ion product of water, equal to 1.0 x 10^-14 at 25 degrees Celsius.
  • Step 4: Find Ka from the given pKa of the conjugate acid (NH4+). Use the formula: Ka = 10^(-pKa).
  • Step 5: Calculate Ka: Ka = 10^(-9.25) = 5.6 x 10^-10.
  • Step 6: Substitute the values of Kw and Ka into the Kb formula: Kb = (1.0 x 10^-14) / (5.6 x 10^-10).
  • Step 7: Perform the division to find Kb: Kb = 1.8 x 10^-5.
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