Calculate pH for 0.10 mole NH3 dissolved in 2L of 0.050 M NH4NO3?

1 Answer
Dec 24, 2015

Answer:

#"pH" = 9.26#

Explanation:

I will assume that you're not familiar with the Henderson - Hasselbalch equation, which allows you to calculate the pH or pOH of a buffer solution.

So, you're interested in finding the pH of a solution that contains #0.10# moles of ammonia, #"NH"_3#, dissolved in #"2 L"# of #"0.050 M"# ammonium nitrate, #"NH"_4"NO"_3# solution.

As you know, ammonia is a weak base, which of course means that id does not ionize completely in aqueous solution to form ammonium ions, #"NH"_4^(+)#, its conjugate acid, and hydroxide ions, #"OH"^(-)#.

Instead, the following equilibrium will be established

#"NH"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(4(aq])^(+) + "OH"_text((aq])^(-)#

Now, you're dissolving the ammonia in a solution that already contains ammonium ions, since ammonium nitrate, a soluble ionic compound, dissociates completely to form

#"NH"_4"NO"_text(3(aq]) -> "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-)#

Notice that ammonium nitrate dissociates in a #1:1# mole ratio with the ammonium ions, which means that

#["NH"_4^(+)] = ["NH"_4"NO"_3] = "0.050 M"#

The concentration of ammonia in this solution will be

#color(blue)(c = n/V)#

#["NH"_3] = "0.10 moles"/"2 L" = "0.050 M"#

Use an ICE table to determine the equilibrium concentration of hydroxide ions in this solution

#" ""NH"_text(3(aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "NH"_text(4(aq])^(+) " "+" " "OH"_text((aq])^(-)#

#color(purple)("I")" " " "0.050" " " " " " " " " " " " " " " " "0.050" " " " " " " " "0#
#color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " "(+x)#
#color(purple)("E")" "(0.050-x)" " " " " " " " " " " "0.050+x" " " " " " "x#

By definition, the base dissociation constant, #K_b#, will be equal to

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

#K_b = ( (0.050+x) * x)/(0.050-x)#

You can find the value for #K_b# here

http://www.bpc.edu/mathscience/chemistry/table_of_weak_bases.html

So, you have

#1.8 * 10^(-5) = ( (0.050+x) * x)/(0.050-x)#

Rearrange this equation to get

#x^2 + (0.050 + 1.8 * 10^(-5)) * x - 9 * 10^(-7) = 0#

This quadratic solution will produce two solutions for #x#, one positive and one negative. Since #x# represents concentration, you can discard the negative value.

You will thus have

#x = 1.799 * 10^(-5)#

This means that you have

#["OH"^(-)] = x = 1.799 * 10^(-5)"M"#

The pOH of the solution will be

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

#"pOH" = - log(1.799 * 10^(-5)) = 4.74#

The pH of the solution will thus be

#color(blue)("pH" = 14 - "pOH")#

#"pH" = 14 - 4.74 = color(green)(9.26)#

SIDE NOTE It is important to notice here that when you have equal concentrations of weak base and conjugate acid, the pOH of the solution will be equal to the #pK_b# of the weak base

#color(blue)(pK_b = - log(K_b))#

#pK_b = -log(1.8 * 10^(-5)) = 4.74#

The Henderson - Hasselbalch equation for weak base / conjugate acid buffers looks like this

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

Notice that when

#["conjugate acid"] = ["weak base"]#

you have

#"pOH" = pK_b + log(1) = pK_b#