Final answer:
The problem involves calculating the final concentrations of all species in a 0.13M KF solution. K+ and F- come directly from the dissociation of KF, while HF and OH- comes from the reaction of F- with water. The concentrations of the four species can be calculated using an ICE table and the equilibrium constant Kb.
Step-by-step explanation:
The original concentration of KF is 0.13 M. KF is a strong electrolyte, so it fully disassociates in water to form K+ and F-. Therefore, [K+] = [F-] = 0.13 M. The fluoride ion (F-) in this solution can react with water to form HF (hydrofluoric acid) and OH-.
Use the Kb value for F- (which is the conjugate base of HF) to calculate the concentrations of HF and OH-. Because the reaction of F- with water is an equilibrium, we can set up an ICE table (Initial, Change, Equilibrium) and use the Kb expression to solve for x, the change in concentration. Ultimately, concentration of HF = concentration of OH- = x, and concentration of F- = 0.13 − x.
In the end, we have four species in the solution: K+, F-, HF, and OH- and we can calculate their concentrations. Please note that the Kb of F- can be calculated from the Ka of HF using the Kw
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