Calculate pH for 0.10 mole NH3 dissolved in 2L of 0.050 M NH4NO3?
1 Answer
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
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,
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
#["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])^(-)#
By definition, the base dissociation constant,
#K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3])#
#K_b = ( (0.050+x) * x)/(0.050-x)#
You can find the value for
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
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
#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#