Question #cb24c

Your strategy here is to use an ICE table to find the equilibrium concentration of the hydronium ions, #"H"_3"O"^(+)#, formed in solution by the partial ionization of formic acid (methanoic acid), #"HCO"_2"H"#, a weak acid.

To get the acid dissociation constant, #K_a#, use the given #pKa#

#K_a = 10^(-pK_a)#

#K_a = 10^(-3.75) = 0.00017778 = 1.78 * 10^(-4)#

So, use an ICE table to get the equilibrium concentration of the hydronium ions

#"CHCO"_2"H"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "CHCO"_text(2(aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#

#color(purple)("I")" " " "0.05" " " " " " " " " " " " " " " "0" " " " " " " " " " " " " " "0#
#color(purple)("C")" "(-x)" " " " " " " " " " " " " "(+x)" " " " " " " " " " "(+x)#
#color(purple)("E")" "0.05-x" " " " " " " " " " " " " "x" " " " " " " " " " " " " " "x#

By definition, the acid dissociation constant will be

#K_a = (["H"_3"O"^(+)] * ["CHO"_2^(-)])/(["CHO"_2"H"])#

#K_a = (x * x)/(0.05 - x) = 1.78 * 10^(-4)#

Since the initial concentration of the acid is relatively small, you cannot use the approximation

#0.05 - x ~~ 0.05#

This means that will have to solve for #x# by using a quadratic equation

#x^2 = 1.78 * 10^(-4) * (0.05 - x)#

#x^2 = 8.9 * 10^(-6) - 1.78 * 10^(-4)x#

#x^2 + 1.78 * 10^(-4)x - 8.9 * 10^(-6) = 0#

This quadratic equation will produce two solutions, one positive and one negative. Since #x# symbolizes concentration, the only solution that will have chemical and physical significance will be the positive one

#x = 0.0028956#

This means that the concentration of hydronium ions will be

#["H"_3"O"^(+)] = x = "0.0028956 M"#

The pH of the solution will be

#"pH" = - log( ["H"_3"O"^(+)])#

#"pH" = - log(0.0028956) = color(green)(2.54)#

SIDE NOTE You can try to solve by using the approximation

#0.05 - x ~~ 0.05#

the pH of the solution will be similar, but the approximation error will be greater than 5%, which indicates that the approximation is not justified.

#0.05 - 0.002983 = 0.04702#

The error is

#|0.04702 - 0.05|/0.05 xx 100 = 5.96%#