Question #9ff52

The idea here is that a weak acid is only partially ionized in aqueous solution, which means that in order to find the solution's concentration of hydronium ions, you must use the acid dissociation constant, #K_a#.

In this case, you have

#"HX"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "X"_ ((aq))^(-)#

By definition, the acid dissociation constant is equal to

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

Now, you can calculate the equilibrium concentrations of the three chemical species by using an ICE table

#" ""HX"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) " "+" " "X"_ ((aq))^(-)#

#color(purple)("I")color(white)(aaaacolor(black)(0.23)aaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0))#
#color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaaaaaaaaacolor(black)((+x))aaaaaaacolor(black)((+x)))#
#color(purple)("E")color(white)(aacolor(black)(0.23-x)aaaaaaaaaaaaaacolor(black)(x)aaaaaaaaaaacolor(black)(x))#

The expression of the acid dissociation constant can thus be written as

#K_a = (x * x)/(0.23 - x)#

which is equivalent to

#1.0 * 10^(-6) = x^2/(0.23 - x)#

Now, because the acid dissociation constant is very small compared to the initial concentration of the acid, you can use the approximation

#0.23 - x ~~ 0.23#

This means that you have

#1.0 * 10^(-6) = x^2/0.23#

Solve for #x# to get

#x = sqrt(0.23 * 1.0 * 10^(-6)) = 0.0004796#

Since #x# represents the equilibrium concentration of hydronium cations, you will have

#["H"_3"O"^(+)] = "0.0004796 M"#

Consequently, the #"pH"# of the solution

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

will be equal to

#"pH" = - log(0.0004796) = color(darkgreen)(ul(color(black)(3.32)))#

The answer is rounded to two decimal places, the number of sig figs you have for the initial concentration of the acid.