Question #bf42d

Start by calculating the number of moles of ammonia present in the initial solution

#125 color(red)(cancel(color(black)("mL"))) * "0.431 moles NH"_3/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.05388 moles NH"_3#

Now, when you add nitric acid, a strong acid that ionizes completely in aqueous solution, to this ammonia solution, the hydronium cations produced by the acid will react with the ammonia to form ammonium cations, #"NH"_4^(+)#, the conjugate acid of ammonia.

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

Notice that the two reactants react in a #1:1# mole ratio; moreover, ammonium cations are produced in #1:1# mole ratios with the two reactants.

In your case, you have fewer moles of hydronium cations than moles of ammonia, which means that the hydronium cations will act as a limiting reagent, i.e. they will be completely consumed before all the moles of ammonia get the chance to react.

After the reaction is complete, the resulting solution will contain

#n_ ("H"_ 3"O"^(+)) = "0 moles" -># completely consumed

#n_ ("NH"_ 3) = "0.05388 moles" - "0.0240 moles" = "0.02988 moles NH"_3#

At the same time, the reaction will produce

#n_ ("NH"_ 4^(+)) = "0 moles + 0.0240 moles = 0.0240 moles NH"_4^(+)#

Calculate the concentrations of ammonia and of ammonium cations in the resulting solution -- we are assuming that the volume did not change upon adding the nitric acid

#["NH"_3] = "0.02988 moles"/(125 * 10^(-3)"L") = "0.239 M"#

#["NH"_4^(+)] = "0.0240 moles"/(125 * 10^(-3)"L") = "0.192 M"#

Now, your solution contains ammonia, a weak base, and ammonium cations, its conjugate acid, in comparable amounts #-># you are dealing with a buffer.

For a weak base/conjugate acid buffer, the Henderson - Hasselbalch equation looks like this

#"pOH" = "p"K_b + log( (["conjugate acid"])/(["weak base"]))#

Look up the #"p"K_b# of ammonia

#"p"K_b = 4.75#

http://www.chembuddy.com/?left=BATE&right=dissociation_constants

You can thus say that you will have

#"pOH" = 4.75 + log( (["NH"_4^(+)])/(["NH"_3]))#

Notice that you have a higher concentration of ammonia than of ammonium cations; this tells you that the #"pOH"# of the solution will be lower than the #"p"K_b# of the weak base.

In other words, the #"pH"# of the solution will be higher than the #"pH"# of the initial ammonia solution.

Plug in your values to find

#"pOH" = 4.75 + log( (0.192 color(red)(cancel(color(black)("M"))))/(0.239color(red)(cancel(color(black)("M")))))#

#"pOH" = 4.65#

As you know, an aqueous solution at room temperature has

#"pH + pOH = 14"#

This means that the #"pH"# of the solution will be equal to

#color(darkgreen)(ul(color(black)("pH" = 14 - 4.65 = 9.35)))#

I have the answer rounded to two decimal places, but you could have it rounded to three decimal places, the number of sig figs you have for your values.