Question #262c6

1 Answer
Nov 3, 2015

Answer:

#"36.2 g NH"_4"Cl"#

Explanation:

Adding ammonium chloride, #"NH"_4"Cl"#, to an ammonia solution will effectively create a buffer solution that contains ammonia, a weak base, and the ammonium ion, #"NH"_4^(+)#, its conjugate acid.

This means that you can use the Henderson - Hasselbalch equation to find the concentration of conjugate base needed to make the pH of the buffer equal to #9#.

#color(blue)(pOH = pK_b + log( (["conjugate acid"])/(["weak base"]))#

To do that, you need to know the value of the base dissociation constant, #K_b#, for ammonia

#K_b = 1.8 * 10^(-5)#

So, use the given pH to find the #"pOH"# of the solution

#"pOH" = 14 - "pH" = 14 - 9.00 = 5.00#

This means that you have

#5.00 = -log(1.8 * 10^(-5)) + log( (["NH"_4^(+)])/(["NH"_3]))#

#5.00 = 4.74 + log( (["NH"_4^(+)])/(["NH"_3]))#

This is equivalent to

#log( (["NH"_4^(+)])/(["NH"_3])) = 0.26#

To get rid of the log, use

# (["NH"_4^(+)])/(["NH"_3]) = 10^(0.26)#

This wil get you

# (["NH"_4^(+)])/(["NH"_3]) = 1.8197#

This tells you tha the ratio between the concentration of conjugate acid and the concentration of weak base must be equal to #1.8197#.

The concentration of ammonium ions will thus be

#["NH"_4^(+)] = 1.8197 * ["NH"_3]#

#["NH"_4^(+)] = 1.8197 * "0.930 M" = "1.6923 M"#

Now, ammonium chloride dissociates completely in aqueous solution to give

#"NH"_4"Cl"_text((aq]) -> "NH"_text(4(aq])^(+) + "Cl"_text((aq])^(-)#

Notice that #1# mole of ammonium chloride produces #1# mole of ammonium ions in solution. This means that the molarity of the ammonium chloride will be equal to that of the ammonium ions

#["NH"_4"Cl"] = ["NH"_4^(+)] = "1.6923 M"#

Assuming that the volume of the solution does not change after adding the ammonium chloride, the number of moles you'd get in the solution will be

#c = n/V implies n = c * V#

#n = "1.6923 M" * 400.0 * 10^(-3)"L" = "0.6769 moles NH"_4"Cl"#

Finally, to get how many grams would contain this many moles, use the compound's molar mass

#0.6769color(red)(cancel(color(black)("moles"))) * " 53.49 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("36.2 g NH"_4"Cl")#

SIDE NOTE It's unreasonable to think that dissolving 36.2 g of ammonium chloride in 400.0 mL of solution will not change the total volume of the buffer - it will!

Now, the thing to remember here is that you need to have that ratio between the concentration of conjugate base and the concentration of weak base regardless of what the total volume of the buffer will be.

This implies that the mole ratio can be used in place of the concentration ratio, since both species are in the same solution.

Here's what I mean. Let's assume that adding the ammonium chloride will increase the volume of the buffer to #0.4 + x# liters.

The Henderson - Hasselbalch equation would then be

#5.00 = 4.74 + log( ("moles of conjugate acid"/"0.4 + x")/("moles of weak base"/"0.4 + x"))#

This will produce the same result

#"moles of ammonium ions"/"moles of ammonia" = 1.8197#

Since you know that your initial ammonia solution contained

#n_"ammonia" = "0.930 M" * 400.0 * 10^(-3)"L" = "0.372 moles"#

it follows that you would still need to add

#n_"ammonium chloride" = 0.372 * 1.8197 = "0.6769 moles"#

of ammonium chloride to get the same buffer pH.

You can test this for various total volumes of the final solution. Let's say that the volume doubles upon the addition of the ammonium chloride. The concentrations of the two species will be

#["NH"_3] = "0.372 moles"/(800.0 * 10^(-3)"L") = "0.465 M"#

#["NH"_4^(+)] = "0.6769 moles"/(800.0 * 10^(-3)"L") = "0.8461 M"#

The #pOH# of the solution will be

#pOH = 4.74 + log(( 0.8461 color(red)(cancel(color(black)("M"))))/(0.465color(red)(cancel(color(black)("M"))))) = 4.99997 ~~5#

So you're good to go.

Now, if they ask you for the new volume, that's a different story. You will need to use the equation

#V_"total" = n_(NH_3) * bar(V)_(NH_3) + n_(NH_4Cl) * bar(V)_(NH_4Cl)#

Here #bar(V)# represents the reciprocal of the molar density

Here's another example to go by

http://socratic.org/questions/what-mass-of-ammonium-chloride-should-be-added-to-2-65-l-of-a-0-160-m-nh3-in-ord