# 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#

The

##### 1 Answer

#### Answer:

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

As you know, an aqueous solution at

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

Notice that the problem gives you the **acid dissociation constant**,

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

Now, if you take **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 mustdecreaseby#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

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

Consequently, the

#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.