Given the following problem:
25.0 mL of 0.30 M hydrochloric acid is added to 40.0 mL of 0.50 M ammonia. Calculate the pH, and concentrations of $\ce{NH3 \text{and} NH4+}$.
To solve such problem you must make some assumptions. Some of which are good, some ok, and some bad. In math $\pi$ has been calculated to 50 trillion digits. Chemistry generally works with 2-4 significant figures.
Note that the concentrations of both hydrochloric acid and ammonia have only been given to 2 significant figures.
Assumptions:
- The final concentrations of $\ce{NH3, NH4+ \text{and} H+/OH-}$ at equilibrium are independent of how the solution was prepared.
The equilibrium doesn't depend on how the solution was mixed. So regardless of how 65 ml of solution containing the equivalent of 7.5 millimoles of hydrochloric acid, and 20 millimoles of ammonia was made, it will have the same equilibrium.
Also $z$ millimoles of ammonium chloride could be used ($z \le 7.5$). Then $7.5-z$ millimoles of hydrochloric acid and $20-x$ moles of ammonia would be needed in the mixture.
- That the volumes of the hydrochloric acid and ammonia solutions are additive.
Certainly good to 2 significant figures, 3 might be good, but 4 is probably an overreach.
- That concentrations can be used instead of activity coefficients.
This is definitely a bad assumption for such concentrated solutions.
- That the reaction $\ce{H+ + NH3 -> NH4+}$ is quantitative.
This is ok for 4 significant figures and the given concentrations.
- Expanding 4, also assume that all the $\ce{NH4+}$ is due to the reaction with hydrochloric acid.
This assumption is good to 2 significant figures, probably 3, but 4 may be a issue. The issue is that there is also the autoionization of water which shifts the equilibrium $\ce{NH3 + H2O <=> NH4+ + OH-}$
$$\ce{H2O <=> H+ + OH-}$$
There is really nothing that can be done with assumptions 2 and 3 without experimental data and/or some sort of functions for the data. So the only thing that can be done is to note the assumptions in the solution.
From Wikipedia the $K_\mathrm{b}$ for ammonium hydroxide is $1.8\times10^{-5}$ ($\mathrm{p}K_\mathrm{b} = 4.7447$, $\mathrm{p}K_\mathrm{a} = 9.2552$).
Solution 1
Ignoring the autoionization of water yields the following charge balance equation:
$\ce{[NH4+] \approx [Cl-]}$
Thus 7.5 millimoles of $\ce{NH4+}$ and 12.5 millimoles of $\ce{NH3}$.
$$\ce{[NH4+] = \dfrac{\pu{7.5 millimoles}}{\pu{65 mL}}} = \pu{0.1153 M}$$
$$\ce{[NH3] = \dfrac{\pu{12.5 millimoles}}{\pu{65 mL}}} =\pu{0.1923 M}$$
Using the hammer known as the Henderson–Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left(\dfrac{\mathrm{[base]}}{\mathrm{[acid]}}\right) = 9.2552 + \log\left(\dfrac{\pu{0.1923 M}}{\pu{0.1153 M}}\right) = 9.4773 $$
Now rounding to two significant figures:
$$\ce{[NH4+]} = \pu{0.12 M}$$
$$\ce{[NH3]} = \pu{0.19 M}$$
$$\mathrm{pH} = 9.47$$
Solution 2 Considering the autoionization of water yields the following charge balance equation:
$\ce{[NH4+] + [H+] = [Cl-] + [OH-]}$
To determine if the first solution is adequate, do two checks using the values from the first solution:
- Is $\ce{[NH4+] \gg [H+]}\ $?
$$\pu{0.12 M}\ \ce{NH4+} \gg \pu{3.4\times10^{-10} M}\ \ce{H+}$$
So this assumption is good.
- Is $\ce{[Cl-] \gg [OH-]}\ $?
$$\pu{0.19 M}\ \ce{Cl-} \gg \pu{3.0\times10^{-5} M}\ \ce{OH-}$$
So this assumption is good to 2 significant figures,
but barely good for 4 significant figures.
Thus the first solution is adequate