The hydrolysis equilibrium constant (K b ) for F - is 1.47E-11 (eq. 1). (Kw=10^-
ID: 528479 • Letter: T
Question
The hydrolysis equilibrium constant (Kb) for F- is 1.47E-11 (eq. 1). (Kw=10^-14)
(eq. 1): F-(aq) + H2O(l) = OH-(aq) + HF(aq)
Calculate the value of the acid dissociation constant (Ka) for HF as shown in the reaction (eq. 2).
(eq. 2): HF(aq) + H2O(l) = H3O+(aq) + F-(aq)
The hydrolysis equilibrium constant (Kb) for F- is 1.47E-11 (eq. 1). (Kw=10^-14)
(eq. 1): F-(aq) + H2O(l) = OH-(aq) + HF(aq)
Calculate the value of the acid dissociation constant (Ka) for HF as shown in the reaction (eq. 2).
(eq. 2): HF(aq) + H2O(l) = H3O+(aq) + F-(aq)
Explanation / Answer
Q1.
find KA for HF
Kw = Ka*KB
(10^-14) = (Ka)(1.47*10^-11)
Ka = (10^-14) /(1.47*10^-11)
Ka= 68027*10^-4
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