How does one calculate the pH of an aqueous solution of a weak acid and a weak base?

Question

In order to create a simple program which calculates titration curves (also to provide some closure in the comments here) I am trying to come up with all the equations so I can calculate the equilibrium proton concentration for any given concentrations of any two given weak acids/bases.

This is a collection of what I've come up with so far and what I assume I still need. If someone could help me back on track I'd kindly appreciate it.

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We're looking at the following reactions: $$\begin{align} \ce{HA &<=>[$K$_{\text{a, 1}}][$K$_{\text{b, 1}}] H+ + A- && (1) }\\ \ce{H+ + B- &<=>[$K$_{\text{b, 2}}][$K$_{\text{a, 2}}] HB && (2)} \end{align}$$ A is the weak acid and B the weak base with the according equilibrium constants (the indices 1 and 2 refer to the acid and base respectively).

For the total masses I've got the following: $$\begin{align} \ce{[A]_{tot} &= [HA] + [A- ] && (3) }\\ \ce{[B]_{tot} &= [HB] + [B- ] && (4) } \end{align}$$

The equilibrium constants are linked via $$ K_\text{a, 1} \times K_\text{b, 2} = \frac{\ce{[A- ][HB]}}{\ce{[HA][B- ]}} $$

Furthermore: $$ \ce{[H+ ]_{eq}} = \frac{\ce{[A]_{tot}} K_\text{a, 1}}{\ce{[A- ]}} - K_\text{a, 1} = \frac{\ce{[HB]}}{K_\text{b, 2} (\ce{[B]_{tot} - [HB]})} $$

For the equilibrium proton concentration I've derived $$ \ce{[H+ ]_{eq}} = K_\text{a, 1} \ce{[HA]} + K_\text{a, 2} \ce{[HB]} + K_\text{w} \ce{[H2O]}$$ where I neglect the influence of the water (the additional term $[\dots] + K_\text{w} \ce{[H2O]}$) since it is so small.

Now, I'm struggling with the following: $$ \ce{[H+ ]_{tot}} = ? $$

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I believe that with the last equation I will be able to solve for $\ce{[H+ ]_{eq}}$. But maybe I'm wrong there as well.

Thanks for any pointers.

Answer 1

Equations and variables

To calculate the proton concentration in a solution as given in the question the following five equations are used for the following five unknown variables: $\ce{[H+ ]}$, $\ce{[HA]}$, $\ce{[A- ]}$, $\ce{[HB]}$ and $\ce{[B- ]}$.

Ion balance: $$ \ce{[H+ ]} = \ce{[A- ] + [B- ]} + \frac{K_\text{w}}{\ce{[H+ ]}} $$

Mass balances: $$\begin{align} \ce{[A]_{tot}} &= \ce{[HA] + [A- ]} \\ \ce{[B]_{tot}} &= \ce{[HB] + [B- ]} \end{align}$$

Law of mass action: $$\begin{align} K_\text{a, 1} &= \frac{\ce{[H+ ][A- ]}}{\ce{[HA]}} \\ K_\text{a, 2} &= \frac{\ce{[H+ ][B- ]}}{\ce{[HB]}} \end{align}$$

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Solution

Combining the above equations we get: $$ \ce{[H+ ]} = \frac{K_\text{a, 1} \ce{[A]_{tot}} }{\ce{[H+ ]}+K_\text{a, 1}} + \frac{K_\text{a, 2} \ce{[B]_{tot}} }{\ce{[H+ ]}+K_\text{a, 2}} + \frac{K_\text{w}}{\ce{[H+ ]} } $$

After some reforming we get a fourth-order polynomial for the proton concentration: $$\begin{align} 0 = & \ce{[H+ ]^4} + \ce{[H+ ]^3}(K_\text{a, 1} + K_\text{a, 2}) \\ &- \ce{[H+ ]^2}(\ce{[A]_{tot}} K_\text{a, 1} + \ce{[B]_{tot}} K_\text{a, 2} - K_\text{a, 1} K_\text{a, 2} + K_\text{w}) \\ & -\ce{[H+ ]}\left[ (\ce{[A]_{tot}} + \ce{[B]_{tot}}) K_\text{a, 1} K_\text{a, 2} + K_\text{w}(K_\text{a, 1} K_\text{a, 2}) \right] \\ & - K_\text{w} K_\text{a, 1} K_\text{a, 2} \end{align}$$

Answer 2

This is the way I would tackle the problem:

$$\begin{align} \ce{HA &<=>[K_\text{a}] H+ + A- && (1) }\\ \ce{B +H2O&<=>[K_\text{b}] BH+ + OH- && (2)} \end{align}$$

The mass balance writes:

$$\begin{align} \ce{[A]_{tot} &= [HA] + [A- ] && (3) }\\ \ce{[B]_{tot} &= [B] + [BH+ ] && (4) } \end{align}$$

The charge balance writes:

$$\begin{align} \ce{[A- ] + [OH- ]&=[H+] + [BH+ ] && (5) } \end{align}$$

Therefore, by introducing eq. (3) into the definition of $K_\text{a}$, one gets:

$$ \ce{[A- ]} = \frac{\ce{[A]_{tot}} K_\text{a}}{\ce{[H+ ]}+K_\text{a}} $$

And using the charge balance (eq. 5), one writes:

$$ \ce{[BH+ ] +[H+ ]}-\frac{K_\text{w}}{\ce{[H+ ] }} = \frac{\ce{[A]_{tot}} K_\text{a}}{\ce{[H+ ]}+K_\text{a}} $$

The same can be done for the base:

$$ \ce{[BH+ ]} = \frac{\ce{[H+ ]}\ce{[B]_{tot}} K_\text{b}}{K_\text{b}\ce{[H+ ] }+K_\text{w}} $$

Putting the last two equations together, one obtains:

$$ \frac{K_\text{w}}{\ce{[H+ ]}} + \frac{\ce{[A]_{tot}} K_\text{a}}{\ce{[H+ ]}+K_\text{a}}=\frac{\ce{[H+ ]}\ce{[B]_{tot}} K_\text{b}}{K_\text{b}\ce{[H+ ] }+K_\text{w}}+\ce{[H+ ]} $$

which is the desired equation with $\ce{[H+ ]}$ as the unique unknown variable.

Hope that helps.

Answer 3

[H+]tot would have to equal [H+]eq. All of your sources of H+ are involved in the equilibria. If you were working with a weak acid and it's conjugate base you could just use the Henderson-Hasselbach equation, eliminating the need for a Kb.

If I understand correctly, this is a titration of a weak acid by a weak base (or vice versa). Therefore H+ would be neutralized immediately, as you add base. So the pH would start low (~pKa, I believe), then slowly increase to where all the acid would be used, and continue to slowly increase until ~pKb. But wow, that endpoint would be hard to find any other way than graphically!