# 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?

##### 1 Answer

This is an unrealistic answer, but I got

By knowing the

#"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

#10^(-"pH") = 10^(-4.88) = ["H"_3"O"^(+)] = x = 1.318 xx 10^(-5)# #"M"#

This gives you a

#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 **completely unrealistic**. The actual

Let's calculate the

#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

#color(red)("pH"_"realistic") = -log(0.001637) = color(red)(2.79)# ,

which is quite a bit less than