# Question #9ff52

##### 1 Answer

#### Answer:

#### Explanation:

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,

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))^(-)#

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

Since

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

Consequently, the

#"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.