# Question #19ceb

##### 2 Answers

Here's what I got.

#### Explanation:

The trick here is to realize that you're dealing with a **buffer solution** that contains ammonia, a weak base, and the ammonium cation, its conjugate acid.

For a weak base/conjugate acid buffer, the

#"pH" = 14 - "p"K_b - log( (["conjugate acid"])/(["weak base"]))#

Since

#"p"K_b = - log(K_b)#

you can say that

#"pH" = 14 + log(K_b) - log( (["conjugate acid"])/(["weak base"]))#

Now, when the buffer contains **equal concentrations** of weak base and conjugate acid, you have

#log( (["conjugate acid"])/(["weak base"])) = log(1) = 0#

and

#"pH" = 14 + log(K_b)#

Notice what happens when you plug in the

#"pH" = 14 + log(1.8 * 10^(-5)) = 9.255#

If you use the fact that your values are rounded to two **sig figs**, you can say that the *decimal places*, so

#"pH" = 9.26#

You can thus say that the final solution contains **equal concentrations** of ammonia and of ammonium cations. Keep this in mind.

Now, ammonia reacts with nitric acid to produce aqueous ammonium nitrate. Since the nitrate anions are *spectator ions*, you can use the **net ionic equation** that describes this neutralization reaction

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

Notice that the two reactants react in a **mole ratio**. This means that in order to completely neutralize the acid, you need to add an **equal number** of moles of ammonia.

The initial solution contained--nitric acid is a **strong, monoprotic acid**, so I'll just write it as

#250 color(red)(cancel(color(black)("mL solution"))) * ("0.050 moles H"_ 3"O"^(+))/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0125 moles H"_3"O"^(+)#

So, in order to neutralize the acid, you need **moles** of ammonia. Notice that every mole of ammonia that consumes **mole** of ammonium cations.

This tells you that a *complete neutralization* will leave the solution with

#"0 moles NH"_3# #"0 moles H"_3"O"^(+)# #"0.0125 moles NH"_4^(+)#

Now, remember that you need to have **equal concentrations** of weak base and conjugate acid. At this point, you have **moles** of ammonium cations, so you need to add **another** **moles** of ammonia to get

#"0.0125 moles NH"_3# #"0.0125 moles NH"_4^(+)#

Mind you, the actual volume of the solution is not important because it is the same for both species. I think that assuming that it remains unchanged, i.e. equal to

The **total number** of moles of ammonia that you added is equal to

#n_ ("NH"_ 3) = overbrace("0.0125 moles")^(color(blue)("to neutralize the acid")) + overbrace("0.0125 moles")^(color(purple)("to get equal concentrations of NH"_3color(white)(.)"and NH"_4^(+)))#

#n_ ("NH"_ 3) = "0.025 moles NH"_3#

Finally, use the **ideal gas law equation**

#color(blue)(ul(color(black)(PV = nRT)))#

where

#P# is the pressure of the gas#V# is the volume it occupies#n# is the number of moles of gas present in the sample#R# is theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof the gas

to find the volume of ammonia that must be bubbled through the initial solution.

Rearrange to sovle for

#PV = nRT implies V = (nRT)/P#

Plug in your values to find--**do not** forget to convert the temperature to *Kelvin* and the pressue to *atm*!

#V = (0.025 color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(740/760color(red)(cancel(color(black)("atm"))))#

#V = "0.628 L"#

Expressed in *milliltiers* and rounded to two **sig figs**, the answer will be

#color(darkgreen)(ul(color(black)(V = "630 mL")))#

You must add 630 mL of ammonia gas.

#### Explanation:

It looks as if you may have an ammonia buffer, so let's calculate the **concentration ratio of base to conjugate acid**.

The chemical equation is

The Henderson-Hasselbalch equation is

So, you have equal concentrations/amounts of

You have neutralized all the

**Calculate the moles of #"HNO"_3#**

**Calculate the moles of #"NH"_3#**

The equation for the neutralization reaction is

The molar ratio of

So, you used 0.0125 mol

**Calculate the volume of #"NH"_3#**

The **Ideal Gas Law** is

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

We can rearrange this formula to get

#V = (nRT)/p#

In this problem,

∴