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

Dec 24, 2015

$\text{pH} = 9.26$

#### 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, ${\text{NH}}_{3}$, dissolved in $\text{2 L}$ of $\text{0.050 M}$ ammonium nitrate, ${\text{NH"_4"NO}}_{3}$ 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, ${\text{NH}}_{4}^{+}$, its conjugate acid, and hydroxide ions, ${\text{OH}}^{-}$.

Instead, the following equilibrium will be established

${\text{NH"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(4(aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

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

${\text{NH"_4"NO"_text(3(aq]) -> "NH"_text(4(aq])^(+) + "NO}}_{\textrm{3 \left(a q\right]}}^{-}$

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

$\textcolor{b l u e}{c = \frac{n}{V}}$

["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

${\text{ ""NH"_text(3(aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "NH"_text(4(aq])^(+) " "+" " "OH}}_{\textrm{\left(a q\right]}}^{-}$

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

${K}_{b} = \left(\left[{\text{NH"_4^(+)] * ["OH"^(-)])/(["NH}}_{3}\right]\right)$

${K}_{b} = \frac{\left(0.050 + x\right) \cdot x}{0.050 - x}$

You can find the value for ${K}_{b}$ here

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

So, you have

$1.8 \cdot {10}^{- 5} = \frac{\left(0.050 + x\right) \cdot x}{0.050 - x}$

Rearrange this equation to get

${x}^{2} + \left(0.050 + 1.8 \cdot {10}^{- 5}\right) \cdot x - 9 \cdot {10}^{- 7} = 0$

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

$x = 1.799 \cdot {10}^{- 5}$

This means that you have

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

The pOH of the solution will be

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

$\text{pOH} = - \log \left(1.799 \cdot {10}^{- 5}\right) = 4.74$

The pH of the solution will thus be

$\textcolor{b l u e}{\text{pH" = 14 - "pOH}}$

$\text{pH} = 14 - 4.74 = \textcolor{g r e e n}{9.26}$

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 $p {K}_{b}$ of the weak base

$\textcolor{b l u e}{p {K}_{b} = - \log \left({K}_{b}\right)}$

$p {K}_{b} = - \log \left(1.8 \cdot {10}^{- 5}\right) = 4.74$

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

$\left[\text{conjugate acid"] = ["weak base}\right]$

you have

$\text{pOH} = p {K}_{b} + \log \left(1\right) = p {K}_{b}$