Question 0c20b

May 7, 2015

Start by writing the balanced chemical equation for the neutralization reaction that takes place between ammonia and hydrochloric acid

$N {H}_{3 \left(a q\right)} + H C {l}_{\left(a q\right)} \to N {H}_{4} C {l}_{\left(a q\right)}$

Notice that you have a $1 : 1$ mole ratio between all the species that take part in the reaction. This will help you determine the number of moles of salt formed in the reaction.

Use the molarities of the two solutions to determine how many moles of each you add together

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{N {H}_{3}} = \text{0.3 M" * 10 * 10^(-3)"L" = "0.003 moles}$ $N {H}_{3}$

${n}_{H C l} = \text{0.1 M" * 10 * 10^(-3)"L" = "0.001 moles}$ $H C l$

Because you have less acid than base, the hydrochloric acid will be consumed completely by the reaction. At the same time, the number of moles of ammonia will decrease by the number of moles of hydrochloric acid that reacted.

Moreover, you'll also produce that many moles of ammonium chloride. This happens because of the aforementioned $1 : 1$ mole ratios - you consume 0.001 moles of $H C l$ and 0.001 moles of $N {H}_{3}$ and produce 0.001 moles of $N {H}_{4} C l$.

Therefore, after the reaction takes place, you'll have

${n}_{H C l} = 0$ $\to$ consumed completely;

n_(NH_3"remaining") = 0.003 - 0.001 = "0.002 moles"# $N {H}_{3}$

${n}_{N {H}_{4} C l} = \text{0.001 moles}$ $N {H}_{4} C l$

The total volume of the solution will be

${V}_{\text{total}} = {V}_{N {H}_{3}} + {V}_{H C l}$

${V}_{\text{total" = 10 + 10 = "20 mL}}$

The new concentrations of the ammonia and ammonium ion (this will be the same as the concentration of the ammonium chloride) will be

$\left[N {H}_{3}\right] = \text{0.002 moles"/(20 * 10^(-3)"L") = "0.1 M}$

$\left[N {H}_{4}^{+}\right] = \text{0.001 moles"/(20 * 10^(-3)"L") = "0.05 M}$

To determine the pH of the solution, you'll need the base dissociation constant, ${K}_{b}$, for ammonia, which is listed at $1.8 \cdot {10}^{- 5}$.

The following equilibrium will be established in solution (use the ICE table method to solve for the concentration of hydroxide ions - more on that here: http://en.wikipedia.org/wiki/RICE_chart)

$\text{ } N {H}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s N {H}_{4 \left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$
I........0.1.......................................0.05.................0
C......(-x)..........................................(+x).................(+x)
E......0.1-x......................................0.05+x..............x

By definition, the base dissociation constant will be equal to

${K}_{b} = \frac{\left[N {H}_{4}^{+}\right] \cdot \left[O {H}^{-}\right]}{\left[N {H}_{3}\right]} = \frac{\left(0.05 + x\right) \cdot x}{0.1 - x} = 1.8 \cdot {10}^{- 5}$

Because ${K}_{b}$ is so small, you can approximate that equation with

$\frac{0.05 \cdot x}{0.1} = 1.8 \cdot {10}^{- 5} \implies x = 3.6 \cdot {10}^{- 5} \text{M}$

Since $x$ is equal to $\left[O {H}^{-}\right]$, the pOH of the solution will be

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left(3.6 \cdot {10}^{- 5}\right) = 4.44$

This means that the pH of the solution will be

$p {H}_{\text{sol}} = 14 - p O H = 14 - 4.44 = \textcolor{g r e e n}{9.56}$