Question

Calculate the `"pH"` of a solution that resulted when `"40 mL"` of a `"0.1-M"` ammonia solution was diluted to `"60 mL"` ?

The `K_a` of the ammonium cation is `5.7 * 10^-10`

Answer 1

`"pH" = 11.0`

Explanation:

Start by calculating the molarity of the diluted ammonia solution.

You know that your stock solution contains

`40 color(red)(cancel(color(black)("mL solution"))) * "0.1 moles NH"_3/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0040 moles NH"_3`

After you dilute this solution by adding enough water to increase its volume from `"40 mL"` to `"60 mL"`, its molarity will be--remember to use the volume of the solution in liters!

`["NH"_3] = "0.0040 moles"/(60 * 10^(-3) quad "L") = "0.0667 M"`

Now, ammonia will act as a weak base in aqueous solution.

`"NH"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-)`

By definition, the expression of the base dissociation constant, `K_b`, is given by

`K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3^(+)])`

As you know, an aqueous solution at `25^@"C"` has

`color(blue)(ul(color(black)(K_a * K_b = 1 * 10^(-14))))`

Notice that the problem gives you the acid dissociation constant, `K_a`, of the ammonium cation, ammonia's conjugate acid. This means that the base dissociation constant for ammonia is equal to

`K_b= (1 * 10^(-14))/(5.7 * 10^(-10)) = 1.75 * 10^(-5)`

Now, if you take `x` `"M"` to be the equilibrium concentration of the ammonium cations and of the hydroxide anions, you can say that, at equilibrium, the solution will also contain

`["NH"_3] = (0.0667 - x) quad "M"`

This basically means that in order for the reaction to produce `x` `"M"` of ammonium cations and `x` `"M"` of hydroxide anions, the concentration of ammonia must decrease by `x` `"M"`.

Plug this back into the expression of the base dissociation constant to find

`1.75 * 10^(-5) = (x * x)/(0.0667 - x)`

`1.75 * 10^(-5) = x^2/(0.0667 - x)`

The value of the base dissociation constant is small enough compared to the initial concentration of the acid to justify the approximation

`0.0667 - x ~~ 0.0667`

This means that you have

`1.75 * 10^(-5) = x^2/0.0667`

which gets you

`x = sqrt(0.0667 * 1.75 * 10^(-5)) = 1.08 * 10^(-3)`

Since `x` `"M"` represents the equilibrium concentration of the hydroxide anions, you can say that your solution contains

`["OH"^(-)] = 1.08 * 10^(-3) quad "M"`

Consequently, the `"pH"` of the solution, which can be found using the fact that at `25^@"C"`, an aqueous solution has

`color(blue)(ul(color(black)("pH + pOH = 14")))`

will be equal to

`"pH" = 14 - "pOH"`

Since

`color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))`

you can say that your solution has

`"pH" = 14 - [- log(1.08 * 10^(-3))] = color(darkgreen)(ul(color(black)(11.0)))`

The answer is rounded to one decimal place, the number of sig figs you have for your values.