Question

Question #19ceb

Answer 1

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"` is equal to

`"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 `K_b` of ammonia

`"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 `"pH"` mus be rounded to two 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 `1:1` 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 `"H"_ 3"O"^(+)`

`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 `0.0125` moles of ammonia. Notice that every mole of ammonia that consumes `1` mole of acid produces `1` 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 `0.0125` moles of ammonium cations, so you need to add another `0.0125` 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 `"250 mL"`, is a fair assumption.

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 the universal gas constant, equal to `0.0821("atm L")/("mol K")`
• `T` is the absolute temperature of the gas

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

Rearrange to sovle for `V`

`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")))`

Answer 2

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

`"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"`

The Henderson-Hasselbalch equation is

`"pOH" = "p"K_text(b) + log((["NH"_4^"+"])/(["NH"_3]))`

`"14.00 - 9.26" = "-log"(1.8 × 10^"-5") + log((["NH"_4^"+"])/(["NH"_3]))`

`4.74 = 4.74 + log((["NH"_4^"+"])/(["NH"_3]))`

`log((["NH"_4^"+"])/(["NH"_3])) = 0.00`

`(["NH"_4^"+"])/(["NH"_3]) = 10^0.00 = 1.00`

So, you have equal concentrations/amounts of `"NH"_4^"+"` and `"NH"_3`.

You have neutralized all the `"HNO"_3` (this formed the `"NH"_4^"+"`) and then added an equal amount of excess `"NH"_3` to form the buffer.

Calculate the moles of `"HNO"_3`

`"Moles of HNO"_3 = 0.250 color(red)(cancel(color(black)("L HNO"_3))) × "0.050 mol HNO"_3/(1 color(red)(cancel(color(black)("L HNO"_3)))) = "0.0125 mol HNO"_3`

Calculate the moles of `"NH"_3`

The equation for the neutralization reaction is

`"NH"_3 + "HNO"_3 → "NH"_4^"+" + "NO"_3^"-"`

The molar ratio of `"NH"_3:"HNO"_3 = 1:1`

So, you used 0.0125 mol `"NH"_3` to neutralize the acid and another 0.0125 mol to form the buffer.

`"Total moles" = "0.025 mol NH"_3`

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,

`n = "0.025 mol"`
`R = "0.082 06 L·atm·K"^"-1""mol"^"-1"`
`T = "(25 + 273.15) K = 298.15 K"`
`p = 740 color(red)(cancel(color(black)("torr"))) × "1 atm"/(760 color(red)(cancel(color(black)("torr")))) = "0.974 atm"`

∴ `V = (0.025 color(red)(cancel(color(black)("mol"))) × "0.082 06 L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K"))))/(0.974 color(red)(cancel(color(black)("atm")))) = "0.63 L" = "630 mL"`