prepare an ICE chart and write the kb equation for F^-
.............Fe^- + HOH == HF + OH^-
initial....0.140...........0......0
change......-x.............x.......x
equil.....0.140-x..........x.......x
Kb for F^- = (Kw/Ka for HF) = (HF)(OH^-)/(HF)
Substitute values from the ICE chart for (HF), (OH^-), and F^-, and for Kw and Ka, solve for x (which is OH^-), convert that to pOH, then to pH.
Determine the Concentration of OH^-1 and pH of a solution that is 0.140 M F^-1.
I did F- +H20 ---> HF + H20
but idk what to solve from there....
1 answer
1 answer