The question was:

Calculate the pH of the solution that results when $\pu{40.0ml}$ of $\pu{0.100M}\ \ce{NH3}$ is mixed with $\pu{20.0ml}$ of $\pu{0.200M}\ \ce{HCl}$ solution.

My thought process was to use the dilution equation $M_1 \times V_1 = M_2 \times V_2$ and then the base dissociation constant expression for ammonia with an ICE table to solve for unknown concentration of either $\ce{OH^-}$ or $\ce{H^+}$.

But if I write out the dissociation equation:

$$\ce{HCl + NH3 <=> NH4+ + Cl^-}$$

Then there is no $\ce{[OH^{-}]}$ or $\ce{[H^+]}$ for me to work with. So then how can I find the pH without that? Is water somehow involved without it being stated? I have solved via an ICE table for the value of $x$; which is the concentration of $\pu{1M}$ of any product or reactant given the stoichiometry, isn’t it? And that value was: $x = 0.001549193$; I did not include my work for that because I felt it was a tangent to the real question at hand.

If I got any terms wrong please forgive me, and if you do want to see my work I can always try and type it in, but I feel without being able to solve for either $\ce{OH^-}$ or $\ce{H^+}$ whats the point?

When I tried to solve for concentrations given in a comment/answer I obtained a weird number, the value of $x = 2.48 \times 10^{-12}$ so then the pH would be 11.60?

I think I figured out my problem, I guess there are different types of ways to write ionic equations: Here was my work from the start.

I needed to determine the concentration of the new dilution, so I used the dilution formula.

$$\begin{align}M_1 \times V_1 &= M_2 \times V_2\\[0.4em] \pu{0.100M} \times \pu{0.040L} &= V_2 \times \pu{0.060M}\\[0.4em] V2 &= \pu{0.6666667M}\end{align}$$

I then wrote the dissociation equation:

$$\ce{HCl + NH3 <=> NH4+ + Cl-}$$

and since $\ce{HCl}$ can dissociate to $\ce{H+ + Cl^-}$ I can keep it in the form $\ce{HCl}$ for calculations on my ICE table.

ICE table

Since $K_\mathrm{a} = 5.6 \times 10^{-10}$ since $\ce{NH4+}$ is the conjugate acid and it has that value.

$$\begin{align}K_\mathrm{a} &= \frac{\text{products}}{\text{reactants}}\\[0.6em] 5.6 \times 10^{-10} &= \frac{[\ce{NH4+}] \times [\ce{Cl-}]}{[\ce{HCl}] \times [\ce{NH3}]}\\[0.6em] 5.6 \times 10^{-10} &= \frac{ [x] \times [x]}{[0.200 - x] \times [0.06667 - x]}\end{align}$$

Now, since $K_\mathrm{a}$ is less than $1 \times 10^{-5}$ we can ignore the $x$ values in the $[0.06667 - x]$ expressions.

$$\begin{align}5.6 \times 10^{-10} &= \frac{ [x]^{2} }{[0.200] \times [0.06667]}\\[0.4em] x &= 0.000002733 = [\ce{H+}]\\[0.4em] \text{pH} &= - \log{0.000002733}\\[0.4em] \text{pH} &= 5.563360368\end{align}$$

Ok so I think I determined that by a fluke, I found a similar problem on Khan academy and since I already wrote such a long solution down I don't really feel like transcribing their attempts.

Using there methods described, I obtained a value for pH of $5.063435995$.

  • 1
    $\begingroup$ Your answer should be to 3 sig figs. $\endgroup$ – Michael F. Jan 13 '19 at 17:44

In aqueous solution, $\ce{HCl}$ dissociates to $\ce{H+}$ and $\ce{Cl-}$

So you have:

$$\ce{H+ + Cl- + NH3 <=> NH4+ + Cl-}$$

and removing the spectator ion $\ce{Cl-}$:

$$\ce{H+ + NH3 <=> NH4+}$$

Then you could write that the equilibrium concentrations are:

$$\begin{align}[\ce{H+}] &= \pu{0.067M} - x\\ [\ce{NH3}] &= \pu{0.067M} - x\\ [\ce{NH4+}] &= x\end{align}$$

But it is easier if you write:

$$\begin{align}[\ce{H+}] &= y\\ [\ce{NH3}] &= y\\ [\ce{NH4+}] &= \pu{0.067M} - y\end{align}$$

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Here is a different approach:

(In order to solve this problem we need to know the $\mathrm{p}K_\mathrm{a}$ of the ammonium ion. In the calculations below I will use $\mathrm{p}K_\mathrm{a}(\ce{NH4+}) = 9.25$).

$$\begin{align}\pu{40.0 ml}\ \pu{0.100 M}\ \text{ammonia} &\Longrightarrow \pu{4.0 mmol}\ \text{ammonia}\\ \pu{20.0 ml}\ \pu{0.200 M}\ \ce{HCl} &\Longrightarrow \pu{4.0 mmol}\ \ce{HCl}\end{align}$$

$\ce{HCl}$ is a strong acid and will convert all the ammonia to ammonium ions. Thus, the end result is the same as dissolving ammonium chloride in water.

Mass balances: $$\begin{align}[\ce{NH4+}] + [\ce{NH3}] &= \frac{\pu{4.0mmol}}{\pu{40.0ml} + \pu{20.0ml}} &&= \pu{0.06667M}\\[0.6em] [\ce{Cl-}] &= \frac{\pu{4.0mmol}}{\pu{40.0ml} + \pu{20.0ml}} &&= \pu{0.06667M}\\[0.6em] [\ce{NH4+}] + [\ce{NH3}] &= [\ce{Cl-}] &&= \pu{0.06667M}\end{align}$$

The equilibrium equation is: $$\frac{[\ce{NH3}]\times[\ce{H3O+}]}{[\ce{NH4+}]} = 10^{-9.25}$$

We also have the proton balance, i.e. number of protons that have been taken up and given off in the reaction: (Initial protolytes: $\ce{H2O}$ and $\ce{NH4+}$): $$[\ce{H3O+}] = [\ce{OH-}] + [\ce{NH3}]$$

Except for very high $\mathrm{pH}$ values, we can always consider $[\ce{NH3}] \gg [\ce{OH-}]$ (this can easily be visualised in a logarithmic diagram of the system). Therefore, at equilibrium we neglect $[\ce{OH-}]$ in comparison with $[\ce{NH3}]$ and we get $[\ce{H3O+}] = [\ce{NH3}]$.

We have: $$\begin{align}[\ce{H3O+}] &= [\ce{NH3}] \tag{1}\\[0.4em] [\ce{NH4+}] + [\ce{NH3}] &= [\ce{Cl-}] = \pu{0.06667 M}\\[0.2em]\Longrightarrow [\ce{NH4+}] &= \pu{0.06667M} - [\ce{H3O+}]\tag{2}\end{align}$$

We insert $(1)$ and $(2)$ in the equilibrium equation above and get:

$$\begin{align}\frac{[\ce{H3O+}]^2}{\pu{0.06667M} - [\ce{H3O+}]} &=10^{-9.25}\\[0.6em] \Longrightarrow [\ce{H3O+}] &= \pu{6.1227e-6M}\\[0.2em] \Longrightarrow \mathrm{pH} &= 5.21\end{align}$$

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