Question

Question #9ff52

Answer 1

`"pH" = 3.32`

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, `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.