Predicting pH of Titrations

How can we predict the $pH$ as we add the titrant? Well, there are two main points that will have a set $pH$, and then two ranges

The first point is the starting $pH$, before adding any titrant. Typically, this can be approximated by assuming the dissociation or association of the acid or the base is negligible. If we do so, we can find $pH$ by using the $pK_a$ or $pK_b$ of the acid or base. Let’s make an ICE table of the situation with Ammonia:

$\begin{matrix} \bold{Condition}&\bold{[NH_3]}&\bold{[NH_4^+]}&\bold{[OH^-]}\\ Initial&X&0&0\\ Change&-x&+x&+x\\ Equilibrium&X&x&x \end{matrix}$

For the equilibrium expression, we drop the $0$, as it isn’t taking part in the reaction.

$K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}$

$K_b=\frac{x\cdot x}X$

$K_b\cdot X=x^2$

$[OH^-]=x$

$pOH = -\log x$

Therefore, we can approximate it as:

$pOH=-\log\sqrt{K_b\cdot X}$

Where $X$ is the initial concentration of the base. This means that pH can be shown to be:

$pH = 14+ \log\sqrt{K_b\cdot X}$

So what’s going on after that? Well, we know the neutralization reaction is occurring then.

$NH_3+HCl\leftrightarrow NH_4^++Cl^-$

So how can we determine the equilibrium of this reaction? First we have to look at the involved reactions, and investigate their equilibrium coefficients.

$NH_3+H^+\leftrightarrow NH_4^+$

$HCl\leftrightarrow H^+ + Cl^-$

As the Hydrochloric acid is a strong acid, the equilibrium for that is assumed to be completely dissociated. Therefore we can dissociate it in the neutralization equation:

$NH_3+H^++Cl^-\leftrightarrow NH_4^++Cl^-$

Now the Chloride ion has become a spectator, and once we eliminate that, we are left with the Ammonia protonation reaction:

$NH_3+H^+\leftrightarrow NH_4^+$

Next, we take a look at the equilibrium equation for this reaction:

$K = \frac{[NH_4^+]}{[NH_3][H^+]}$

This looks mighty similar to the $K_a$ equation for Ammonium:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

As the two equations are merely the inverse of one another, we can look at this part in terms of $K_a$.

As we are tracking $pH$, we need to use the Henderson-Hasselbalch equation:

$pH=pK_a+\log\frac{[A^-]}{[HA]}$

Now applying that to our situation, we have:

$pH=pK_a+\log\frac{[NH_3]}{[NH_4^+]}$

We need to figure out how to track $[NH_3]\text{ and }[NH_4^+]$. If we assume that there is negligible protonation of ammonia before we started adding the acid, we can safely say that the protonation of the Ammonia is solely from the addition of $[H^+]$ from the acid. Here’s the neutralization equation again:

$NH_3+HCl\leftrightarrow NH_4^++Cl^-$

On a mole basis, the initial situation is approximately the following:

$n_{NH_3}=C_{NH_3}V_{NH_3}$

And as we add the acid, we can determine the moles acid as:

$n_{HCl}=C_{HCl}V_{HCl}$

We don’t know the exact number of moles of Ammonium, as we don’t know the concentration (that’s the whole point of the titration!) We do know every other variable for these equations, as we are adding known amounts of a known concentration of $HCl$ to a known volume of the Ammonia solution.

We know that one mole of $HCl$ protonates one mole of $NH_3$. Therefore, we can track the moles of Ammonia in solution by subtracting the moles of $HCl$ we have added. Similarly, one mole of Ammonium is generated by one mole of $HCl$ addition, so that can be tracked the same.

$n_{NH_3}=n_{NH_3, start}-n_{HCl\ added}$

$n_{NH_4^+}=n_{HCl\ added}$

Now that we have the moles, we need to re-write the Henderson-Hasselbalch equation to utilize the moles instead of the concentration.

$pH=pK_a+\log\frac{[NH_3]}{[NH_4^+]}$

Recall:

$[A]=\frac{n_A}{V_{solution}}$

Throughout the titration, $V_{solution}$ is going to be the sum of the initial volume and the volume of the titrant added:

$V_{solution}=V_{analyte}+V_{titrant}$

$pH = pK_a+\log\frac{\frac{n_{NH_3}}{V_{solution}}}{\frac{n_{NH_4^+}}{V_{solution}}}$

$pH=pK_a+\log\frac{n_{NH_3}}{n_{NH_4^+}}$

$pH=pK_a+\log\frac{n_{NH_3, start}-n_{HCl\ added}}{n_{HCl\ added}}$

The overall goal is to plot $pH\text{ vs. }V_{titrant}$, so we need to get this equation into that form. To do so, we replace the equations for the moles of each $HCl\text{ and }NH_3$.

$n_{HCl\ added} = C_{HCl}V_{titrant}$

$n_{NH_3}=C_{NH_3}V_{analyte}$

$pH=pK_a+\log\frac{C_{NH_3}V_{analyte}-C_{HCl}V_{titrant}}{C_{HCl}V_{titrant}}$

Of course, we don’t know the concentration of ammonia, so we can’t just draw up the plot. We need to track $pH$ as we add the $HCl$.

What this shows is that the $pH$ of the solution will start to drop considerably once the amount of $HCl$ added gets very close to the moles of $NH_3$, as the $\frac{n_{NH_3, start}-n_{HCl\ added}}{n_{HCl\ added}}$ term will approach $0$. Therefore, you expect to see a drop in $pH$ as it reaches the equivalence point. At the equivalence point, however, the equation falls apart, as $n_{NH_3}-n_{HCl\ added}=0$, just by the definition of the equivalence point. If we want to predict the $pH$ at the equivalence point, we have to use a different approach. To find it, we need to start by making a couple assumptions.

First, being at the equivalence point, we can say that the moles of $HCl$ added is equivalent to the moles of $NH_3$ initially present.

$n_{HCl, added}=n_{NH_3,\ initial}$

$C_{NH_3}V_{NH_3}=C_{HCl}V_{HCl}$

Second, we need to assume that because the neutralization reaction has completely progressed, meaning that all of the $NH_3$ has been converted to $NH_4^+$.

$NH_3+HCl\leftrightarrow NH_4^++Cl^-$

This implies that the number of moles of $NH_4^+$ at the equivalence point is equal to the number of moles of $HCl$ added.

$n_{HCl, added}=n_{NH_3,\ initial}=n_{NH_4^+,\ at\ equivalence}$

Now let’s look at the acid equilibrium for Ammonium.

$NH_4^+\leftrightarrow NH_3+H^+$

We can use this equation to solve for $pH$. The ICE table for this reaction is:

$\begin{matrix} \bold{Condition}&\bold{[NH_4^+]}&\bold{[NH_3]}&\bold{[H^+]}\\ Initial&[NH_4^+]&0&0\\ Change&-x&+x&+x\\ Equilibrium&[NH_4^+]&x&x \end{matrix}$

We need to come up with a value for $[NH_4^+]$. We know the number of moles:

$C_{HCl}V_{HCl}=n_{HCl,\ added}=n_{NH_4^+}$

We also know the volume of the solution:

$V=V_{analyte}+V_{titrant}=V_{NH_3}+V_{HCl}$

Substituting these values gives us:

$[NH_4^+]=\frac{n_{NH_3}}{V_{solution}}$
$[NH_4^+]=\frac{C_{HCl}V_{HCl}}{V_{NH_3}+V_{HCl}}$

We can actually get rid of the volumetric variables in this equation, as the equivalence point’s $pH$ is independent of the volumes, and solely reliant on the concentrations involved. To do this, we need to go back to the equivalence equation:

$C_{NH_3}V_{NH_3}=C_{HCl}V_{HCl}$

Now solve for $V_{NH_3}$

$V_{NH_3}=\frac{C_{HCl}V_{HCl}}{C_{NH_3}}$

Plug this into the $[NH_4^+]$ equation:

$[NH_4^+]=\frac{C_{HCl}V_{HCl}}{\frac{C_{HCl}V_{HCl}}{C_{NH_3}}+V_{HCl}}$

Now we can cancel the $V_{HCl}$.

$[NH_4^+]=\frac{C_{HCl}V_{HCl}}{\left ( \frac{C_{HCl}}{C_{NH_3}}+1\right)V_{HCl}}$

$[NH_4^+]=\frac{C_{HCl}}{\frac{C_{HCl}}{C_{NH_3}}+1}$

That looks bad, so let’s simplify it.

$[NH_4^+]=\frac{C_{HCl}}{\frac{C_{HCl}}{C_{NH_3}}+1}\cdot\frac{C_{NH_3}}{C_{NH_3}}$

$[NH_4^+]=\frac{C_{HCl}C_{NH_3}}{C_{HCl}+C_{NH_3}}$

Now we can plug this into the equilibrium equation.

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

$K_a[NH_4^+]=[NH_3][H^+]$

$K_a\frac{C_{HCl}C_{NH_3}}{C_{HCl}+C_{NH_3}}=x\cdot x$

Use this to solve for $pH$.

$pH = -\log [H^+]$

$pH = -\log x$

$pH=-\log\sqrt{K_a\frac{C_{HCl}C_{NH_3}}{C_{HCl}+C_{NH_3}}}$

That $K_a$ is a bit confusing, as it’s the $K_a$ of $NH_4^+$, the conjugate acid of the base we are titrating. Perhaps it would be easier to look at this solely in terms of $HCl$ and $NH_3$. To do this, let’s use the relationship between $K_a$ and $K_b$.

$pK_a+pK_b=pK_w$

$-\log K_a -\log K_b=-\log K_w$

$-\log\left(K_a\cdot K_b\right)=-\log K_w$

$K_a\cdot K_b=K_w$

$K_a=\frac{K_w}{K_b}$

Now we can find $pH$ as:

$pH=-\log\sqrt{\frac{K_w}{K_b}\cdot \frac{C_{HCl}C_{NH_3}}{C_{HCl}+C_{NH_3}}}$

Now we are using the $K_b$ of our starting base, which makes things a bit clearer.

One more part of the titration is after the equivalence point. Generally, a titration will stop once the equivalence point it reached, but sometimes you want to keep going, as you wish to examine the whole trek of the $pH$, in order to find the equivalence point as the center of the $pH$ drop.

So if we want to track the $pH$ as we add even more acid, we have to realize that by this point, all of the $NH_3$ has been converted. Therefore, all the $HCl$ we add from this point on will have no $NH_3$ left to protonate, so it will contribute directly to the $pH$ by simply dissociating (remember it is a strong acid, so we assume complete dissociation).

For this leg of the titration, we will assume that the $NH_4^+$ no longer has a significant impact on the $pH$, and the $[H^+]$ instead comes from the unreacted $HCl$ that is added.

To track the $pH$, we will assume complete dissociation of $HCl$:

$HCl\leftrightarrow H^++Cl^-$

$\begin{matrix} \bold{Condition}&\bold{[HCl]}&\bold{[H^+]}&\bold{[Cl^-]}\\ Initial&[HCl]&0&0\\ Change & -x&+x&+x\\ Equilibrium&[HCl]-x&x&x \end{matrix}$

So what is the concentration of $[HCl]$? Well, let’s figure that out based off a molar basis. Recall that as it was first added, the $HCl$ was reacting with $NH_3$:

$HCl+NH_3\leftrightarrow NH_4^++Cl^-$

Therefore, the unreacted $HCl$ must be what is left after fully protonating the $NH_3$:

$n_{HCl}=n_{HCl,\ added}-n_{NH_3,\ initial}$

Now we can define $n_{HCl}$ in our usual terms:

$n_{HCl}=C_{HCl}V_{HCl}-C_{NH_3}V_{NH_3}$

Use this to determine $[HCl]$.

$[HCl]=\frac{n_{HCl}}{V_{solution}}$

$[HCl]=\frac{n_{HCl}}{V_{HCl}+V_{NH_3}}$

$[HCl]=\frac{C_{HCl}V_{HCl}-C_{NH_3}V_{NH_3}}{V_{HCl}+V_{NH_3}}$

$\begin{matrix} \bold{Condition}&\bold{[HCl]}&\bold{[H^+]}&\bold{[Cl^-]}\\ Initial&[HCl]&0&0\\ Change & -x&+x&+x\\ Equilibrium&\frac{C_{HCl}V_{HCl}-C_{NH_3}V_{NH_3}}{V_{HCl}+V_{NH_3}}-x&x&x \end{matrix}$

We are assuming completed dissolution, so:

$\frac{C_{HCl}V_{HCl}-C_{NH_3}V_{NH_3}}{V_{HCl}+V_{NH_3}}-x=0$

$x=\frac{C_{HCl}V_{HCl}-C_{NH_3}V_{NH_3}}{V_{HCl}+V_{NH_3}}$

Now we can find $pH$:

$pH = -\log[H^+]$

$pH=-\log x$

$pH = -\log \frac{C_{HCl}V_{HCl}-C_{NH_3}V_{NH_3}}{V_{HCl}+V_{NH_3}}$

With these equations, we can successfully predict the $pH$ at any point of a titration, though it requires knowing the initial concentration of the Ammonia. We could use a program to fit these equations to a measured $pH$ graph, but that is quite a bit more complicated than predicting the $pH$ from these equations.