# What is the #"pH"# of a #"1.5-M"# solution of ammonia? The dissociation constant of ammonia at #25.0^@"C"# is #1.80 * 10^-5#.

##### 1 Answer

#### Answer:

#### Explanation:

As you know, ammonia acts as a **weak base** in aqueous solution, so right from the start, you should expect the

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

The ratio that exists between the **equilibrium concentrations** of the ammonium cations and of the hydroxide anions and the **equilibrium concentration** of ammonia is given by the **base dissociation constant**,

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

Now, ammonia will only *partially ionize* to produce ammonium cations and hydroxide anions. If you take **that ionizes**, you can say that, at equilibrium, the solution will contain

#["NH"_4^(+)] = ["OH"^(-)] = x quad "M"# This happens because

every moleof ammoniathat ionizesproduces#1# moleof ammonium cations and#1# moleof hydroxide anions.So if

#x# #"M"# ionizes, you can expect the solution to contain#x# #"M"# of the two ions.

The solution will also contain

#["NH"_3] = (1.5 - x) quad "M"# When

#x# #"M"# ionizes, the initial concentration of ammonia willdecreaseby#x# #"M"# .

This means that the expression of the base dissociation constant will now take the form

#K_b = (x * x)/(1.5 - x)#

which is equal to

#1.80 * 10^(-5) = x^2/(1.5 - x)#

Notice that the value of the base dissociation constant is *significantly smaller* than the initial concentration of the base. This tells you that you can use the approximation

#1.5 -x ~~ 1.5#

because the concentration of ammonia **that ionizes** will be significantly lower than the initial concentration of the base, i.e. the ionization equilibrium will **lie to the left**.

You now have

#1.80 * 10^(-5) = x^2/1.5#

Rearrange and solve for

#x = sqrt(1.5 * 1.80 * 10^(-5)) = 0.005196#

Since

#["OH"^(-)] = "0.005196 M"#

Now, an aqueous solution at

#"pH + pOH = 14"#

Since

#"pOH" = - log(["OH"^(-)])#

you can say that the

#"pH" = 14 + log(["OH"^(-)])#

Plug in your value to find

#"pH" = 14 + log(0.005196) = color(darkgreen)(ul(color(black)(11.72)))#

The answer is rounded to two **decimal places** because you have two **sig figs** for the molarity of the solution.