Answer
NaF
Work Step by Step
Given pH, the pOH can be calculated with 14-8.08= 5.92
The [OH-] can be calculated with $10^{-5.92}$= $1.2\times$$10^{-6}$ M
The molarity of the unknown salt solution is given by $0.050 mol\div$0.500 L= 0.10 M.
Set up an ICE table, and you will find that $K_{b}$ = ($1.2\times$$10^{-6}$)^2 divided by (0.1-$1.2\times$$10^{-6}$)= $1.4\times$$10^{-11}$
Use this to find $K_{a}$ by doing $($Kw$)\div$($K_{b}$) where $K_{w}$= $1\times$$10^{-14}$. $K_{a}$= $6.9\times$$10^{-4}$, which matches the $K_{a}$ value of HF, meaning the $K_{b}$ matches that of the conjugate base $F^{-}$.
