And for your problem:
#5.17=4.069+log_(10)([[C_3H_7O_2^(-)]]/[[C_3H_7CO_2H]])#
Clearly, #log_(10)([[C_3H_7O_2^(-)]]/[[C_3H_7CO_2H]])=5.17-4.069=1.10#
And we take antilogs.........
#10^(1.10)=[[C_3H_7O_2^(-)]]/[[C_3H_7CO_2H]]=12.62#
But by spec. #[C_3H_7CO_2H]=0.50*mol*L^-1#.
So #[C_3H_7O_2^-]=6.31*mol*L^-1#
For convenience, we would first prepare a #6.81*mol*L^-1# solution of #"propionic acid"# (which as I recall would not have a very pleasant smell!), and then add a #6.31*mol# #KOH# to a #1*L# volume.
Alternatively, we could add #6.31*molxx96.07*g*mol^-1=606.1*g# #"sodium propionate"# to a #1*L# volume of #0.500*mol*L^-1# #"propionic acid"#. (The volumes would not change too much!)
