If Kb for NH₃ is 1.8 × 10⁻⁵ at 25 0C, what is the pKa of its conjugate acid (NH₄⁺)?

Question

BlackTom AI · Accepted Answer

The task presents a question about acid-base equilibria involving NH3 and its conjugate acid NH4+. 
First, recall the relationship between pKa and pKb for conjugate acid-base pairs at 25°C: pKa + pKb = pKw, where pKw ≈ 14.00. 
Next, determine pKb from the given Kb for NH3: Kb = 1.8 × 10^−5. 
Calculate pKb as the negative logarithm: pKb = −log10(1.8 × 10^−5) ≈ 4.74. 
Using the pKw relation, compute pKa for NH4+: pKa = pKw − pKb ≈ 14.00 − 4.74 = 9.26. 
Rounding considerations: depending on the significant figures, this value is typically given as about 9.26 or 9.25. 
Now, regarding the answer options: the provided dataset shows an empty set of answer choices, so there are no individual alternatives to evaluate or compare. 
Despite the lack of listed options, the derived pKa value for the conjugate acid NH4+ based on the given Kb of NH3 is approximately 9.26, which aligns with common rounded values near 9.25 to 9.26. 
In practical terms, if you were selecting from options, you would choose the one closest to 9.26; any option far from this value would be incorrect due to deviation in the pKa calculation stemming from the pKw and the logarithmic conversion.

CHM-1020-LD01.2025SU Exam 2

If Kb for NH₃ is 1.8 × 10⁻⁵ at 25 0C, what is the pKa of its conjugate acid (NH₄⁺)?

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