The pH of 0.02 M solution of an unknown weak acid is 3.7, how would you find the pka of this acid?
1 Answer
Knowing the pH, you know the concentration of protons:
#-log["H"^(+)] = "pH" = 3.7#
#["H"^(+)] = 10^(-3.7)# #"M"#
Now, since the weak (monoprotic) acid dissociates into its conjugate base and a proton, the
#"HA" rightleftharpoons "A"^(-) + "H"^(+)#
Hence,
#K_a = (["H"^(+)]_"eq"^2)/(["HA"]_"eq"]#
Don't forget that the
#= (["H"^(+)]_"eq"^2)/(["HA"]_i - ["H"^(+)]_"eq")#
#= (10^(-3.7) "M")^2/(0.02 "M" - 10^(-3.7) "M")#
#K_a = 2.0105xx10^(-6)# #"M"#
Thus:
#color(blue)("pKa") = -log("K"_a)#
#= -log(2.01059xx10^-6)#
# ~~ color(blue)(5.70)# where the logarithm of any number is unitless.