# Question 0d9d2

Apr 13, 2015

The concentrations are [NH₄⁺] = 0.45 mol/L; [Cl⁻] = 0.45 mol/L; [NH₃] = 0.15 mol/L; [OH⁻] = 5.93 × 10⁻⁶ mol/L; [H₃O⁺] = 1.69 × 10⁻⁹ mol/L.

Step 1. Identify the species.

You have two equilibria here — the ionization of ammonia and the ionization of water.

The equations are:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
2H₂O = H₃O⁺ + OH⁻

The species involved in the equilibria are NH₃, NH₄⁺, OH⁻, and H₃O⁺.

You also have the Cl⁻ as a spectator ion from the NH₄Cl (5 species in total).

Step 2. Determine the concentrations.

You know immediately that

(1) [NH₃] = 0.15 mol/L

(2) [NH₄⁺] = 0.45 mol/L

(3) [Cl⁻] = o.45 mol/L

Since this is a buffer, we can use the Henderson-Hasselbalch Equation:

"pOH" = "p"K_"b" + log((["NH"_4^+])/(["NH"_3])) = 4.75 + log((0.45 cancel("mol/L"))/(0.15 cancel("mol/L")))

$= 4.75 + 0.48 = 5.23$

["OH"^(-)] = 10^"-pOH" = 10^-5.23

(4) ["OH"^(-)] = 5.93 × 10^-6 "mol/L"

["H"_3"O"^+] = K_w/(["OH"^(-)]) = (1.00 × 10^-14)/(5.93 × 10^-6)

(5) ["H"_3"O"^+] = 1.69 × 10^-9 "mol/L"#

Apr 13, 2015

$\left[N {H}_{3 \left(a q\right)}\right] = 0.15 \text{mol/l}$

$\left[N {H}_{4 \left(a q\right)}^{+}\right] = 0.45 \text{mol/l}$

$\left[C {l}_{\left(a q\right)}^{-}\right] = 0.45 \text{mol/l}$

$\left[{H}_{\left(a q\right)}^{+}\right] = 1.69 \times {10}^{- 9} \text{mol/l}$

$\left[O {H}_{\left(a q\right)}^{-}\right] = 5.93 \times {10}^{- 6} \text{mol/l}$

The mixture contains a lot of unreacted ammonia molecules and lots of ammonium ions.

Ammonium ions are slightly acidic:

$N {H}_{4 \left(a q\right)}^{+} r i g h t \le f t h a r p \infty n s N {H}_{3 \left(a q\right)} + {H}_{\left(a q\right)}^{+}$

We can make some assumptions which means we don't have to go into working out equilibrium concentrations.

We can write an expression for ${K}_{a} \Rightarrow$

${K}_{a} = \frac{\left[N {H}_{3}\right] \left[{H}^{+}\right]}{\left[N {H}_{4}^{+}\right]} = 5.62 \times {10}^{- 10} \text{mol/l}$

The presence of ammonia in the mixture forces the equilibrium to the left.

This means we can assume the ammonium ion concentration is what we started with - $0.45 \text{M}$ and that the ammonia concentration is all due to the initial ammonia - $0.15 \text{M}$

Putting in the numbers:

$5.62 \times {10}^{- 10} = \frac{0.15 \times \left[{H}^{+}\right]}{0.45}$

From which:

$\left[{H}^{+}\right] = 1.69 \times {10}^{- 9} \text{mol/l}$

To find $\left[O {H}^{-}\right]$ we use the ionic product of water:

$\left[{H}^{+}\right] \left[O {H}^{-}\right] = {10}^{- 14} m o {l}^{2} . {l}^{- 2} \text{@} 298 K$

$\left[O {H}^{-}\right] = \frac{{10}^{- 14}}{1.69 \times {10}^{- 9}} = 5.93 \times {10}^{- 6} \text{mol/l}$

The chloride ions take no part in any of this so their concentrations remain at $0.45 \text{mol/l}$.