Open Question
Given that Ka for HIO is 3.2×10−11 at 25 °C, what is the value of 𝐾b for IO− at 25 °C?
17. Acid and Base Equilibrium
Ka and Kb
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If the Kb of NH3 is 1.76 x 10-5, determine the acid dissociation constant of its conjugate acid.
A
1.08 x 10-10
B
5.68 x 10-10
C
5.68 x 10-9
D
1.0 x 10-14
Verified step by step guidance1
Identify the relationship between the base dissociation constant (Kb) and the acid dissociation constant (Ka) for a conjugate acid-base pair. This relationship is given by the equation: Ka * Kb = Kw, where Kw is the ion-product constant of water, typically 1.0 x 10^-14 at 25°C.
Given that the Kb of NH3 is 1.76 x 10^-5, and knowing that NH3 is the base, its conjugate acid is NH4+.
Use the relationship Ka * Kb = Kw to solve for Ka, the acid dissociation constant of NH4+. Rearrange the equation to find Ka: Ka = Kw / Kb.
Substitute the known values into the equation: Ka = (1.0 x 10^-14) / (1.76 x 10^-5).
Perform the division to find the value of Ka, which will give you the acid dissociation constant of the conjugate acid NH4+.
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