When #"0.2 M"# of propionic acid dissolves in water, the #"pH"# was found to be #4.88#. Find the #K_a# of propionic acid?

Question

Natalie Anton · Accepted Answer

This is an unrealistic answer, but I got #8.69 xx 10^(-10)#. The actual #K_a# is very different, at #1.34 xx 10^(-5)#!! The #"pH"# given was unrealistic, so the #K_a# calculated was unrealistic. By knowing the #"pH"#, you know the equilibrium concentration of #"H"^(+)#, which means you know what #x# is. #"PropCOOH"(aq) + "H"_2"O"(l) rightleftharpoons "PropCOO"^(-)(aq) + "H"_3"O"^(+)(aq)# #"I"" ""0.2 M"" "" "" "" "-" "" "" ""0 M"" "" "" "" "" ""0 M"#
#"C"" " - x" "" "" "" "-" "" "" "+x" "" "" "" "" "+x#
#"E"" "(0.2 - x) "M"" "-" "" "" "" "x" M"" "" "" "" "x" M"# Since #"pH" = -log["H"_3"O"^(+)]#: #10^(-"pH") = 10^(-4.88) = ["H"_3"O"^(+)] = x = 1.318 xx 10^(-5)# #"M"# This gives you a #K_a# expression of: #color(blue)(K_a) = (["PropCOO"^(-)]["H"_3"O"^(+)])/(["PropCOOH"])# #= x^2/(0.2 - x)# #= (1.318 xx 10^(-5))^2/(0.2 - 1.318 xx 10^(-5))# #= color(blue)(8.69 xx 10^(-10))# By the way, this #K_a#, although calculated correctly, is completely unrealistic. The actual #K_a# of propionic acid is around #1.34 xx 10^(-5)#... so the #"pH"# is unrealistic. Let's calculate the #"pH"# using the correct #K_a# and compare. Since #K_a# is small (on the order of #10^(-5)# or less), #K_a ~~ x^2/(0.2)# #=> x ~~ sqrt(0.2K_a) = sqrt(0.2 * 1.34 xx 10^(-5)) = "0.001637 M"#  As a result, a more realistic #"pH"# would have been: #color(red)("pH"_"realistic") = -log(0.001637) = color(red)(2.79)#, which is quite a bit less than #4.88#...

Avery Adams · Answer

undefined To find the Ka of propionic acid, you can use the formula:

Ka = [H+][C3H5O2-] / [C3H6O2]

Given that the pH is 4.88, you can find the concentration of H+ ions. Then, since propionic acid is a weak acid, you can assume that the concentration of propionic acid (C3H6O2) is equal to the initial concentration of propionic acid (0.2 M) minus the concentration of H+. Finally, plug the values into the Ka formula to solve for Ka.