posted by juan777 .
if 8.7 g of butanoic acid is dissolved in enough water to make 1.0 L of solution, what is the resulting pH
Let's call butanoic acid HB. Then
HB ==> H^+ + B^-
K = (H^+)(B^-)/(HB)
Convert 8.7 g butanoic acid to moles and that dissolved in 1.0L = the molarity.
You don't list K but you must have it. Substitute K and set up an ICE chart for the equilibrium, solve for H^+ and convert to pH.