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What is the Kb of ammonia (NH3) if the pKa of its conjugate acid (NH4+) is 9.25?
What is the Kb of ammonia (NH3) if the pKa of its conjugate acid (NH4+) is 9.25?
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Practice Questions
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Q1
What is the Kb of ammonia (NH3) if the pKa of its conjugate acid (NH4+) is 9.25?
1.8 x 10^-5
5.6 x 10^-10
1.0 x 10^-14
3.2 x 10^-5
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Kb = Kw / Ka = 1.0 x 10^-14 / 5.6 x 10^-10 = 1.8 x 10^-5
Questions & Step-by-step Solutions
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Q
Q: What is the Kb of ammonia (NH3) if the pKa of its conjugate acid (NH4+) is 9.25?
Solution:
Kb = Kw / Ka = 1.0 x 10^-14 / 5.6 x 10^-10 = 1.8 x 10^-5
Steps: 7
Show Steps
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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