What is the #"pH"# of a #"0.20-M"# solution of formic acid?
#pK_a = 3.75#
1 Answer
For part (a):
Explanation:
I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.
So, you're dealing with formic acid,
You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes
#"HCOOH"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons" " "HCOO"_text((aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#
You need to use the acid's
#color(blue)(K_a = 10^(-pK_a))#
In your case, you have
#K_a = 10^(-3.75) = 1.78 * 10^(-4)#
You know that the equilibrium constant is also equal to
#K_a = (["HCOO"^(-)] * ["H"_3"O"^(+)])/(["HCOOH"])#
#K_a = (x * x)/(0.20 - x)#
Since
#K_a = x^2/0.2 impliesx = sqrt(0.2 * 1.78 * 10^(-4)) = 5.97 * 10^(-3)#
The concentration of the hydronium ions will thus be
#x = ["H"_3"O"^(+)] = 5.97 * 10^(-3)"M"#
This means that the solution's pH will be
#pH_"sol" = -log(["H"_3"O"^(+)])#
#pH_"sol" = -log(5.97 * 10^(-3)) = color(green)(2.22)#