I'm an industrial pharmacist needing some help with some chemistry basics.

I'm trying to calculate the theoretical pH of an effervescent preparation containing 2 mmol citric acid ($\ce{H3C6H5O7}$), 4 mmol sodium bicarbonate ($\ce{NaHCO3}$), and 0.4 mmol sodium carbonate ($\ce{Na2CO3}$). These powders are dissolved in 100 ml of pH 6.0 water.

Citric acid: $\mathrm{p}K_\mathrm{a1}$ = 3.13, $\mathrm{p}K_\mathrm{a2}$ = 4.76, $\mathrm{p}K_\mathrm{a3}$ = 6.39
Sodium bicarbonate: $\mathrm{p}K_\mathrm{a}$ = 10.329, 6.351 (carbonic acid)
Sodium carbonate: $\mathrm{p}K_\mathrm{b}$ = 3.67

I don't know how to calculate pH in a case like this, where both acid and base are weak and polyprotic. If someone can do an example calculation, I'd really appreciate it.

  • 3
    $\begingroup$ Please try to have a go at solving a problem, post your attempt to solve it as part of the question. Stack Exchange is like life, you get out of it what you put into it. Keep in mind that it is not a homework answer machine, if you can show us what you have tried to do so far you will be likely to get more help here. $\endgroup$ Commented Jun 26, 2018 at 4:11
  • $\begingroup$ How accurate do you really need to be? $\endgroup$
    – ringo
    Commented Jun 26, 2018 at 12:17
  • $\begingroup$ I don't need it to be very accurate. I just want to understand the procedure to determine the pH (and possibly the concentration of other ions, the equilibrium of the reaction) when given the amount and pKa/pKb of reactants. $\endgroup$
    – Gnad
    Commented Jun 26, 2018 at 12:37

2 Answers 2


You have a system of an acid, an ampholyte and a base. As you are interested in the pH value, your main question is what does "produce" or "consume" them. The proton concentration for your specific system is (citric Acid, sodium Bicarbonate, sodium Carbonate): $$\ce{[H+] = [OH-] + \underbrace{[H2A-] + 2[HA^2-] + 3[A^3-]}_{citric acid's bases} \underbrace{+ [B^2-] - [H2B]}_{bicarbonate's deprotonated\\ and protonated form} \underbrace{- [CH+] - 2 [CH2^2+]}_{carbonate's sinlgy and\\ doubly protonated form}}$$

The multipliers arise from the fact, that a n-times deprotonated/protonated ion yields/consumes n protons.

Now the question is, where you get these concentrations from. It is by solving a linear equation system for each subsystem.

Citric acid, to be deprotonated three times: \begin{align} \ce{[H3A] + [H2O] &<=> [H2A-] + [H3O+]\\ [H2A-] + [H2O] &<=> [HA^2-] + [H3O+]\\ [HA^2-] + [H2O] &<=> [A^3-] + [H3O+]} \end{align} This gives you the following linear equation system \begin{align} k_{a1} &= \ce{\frac{[H2A-][H3O+]}{[H3A]}} \\ k_{a1} \ce{[H3A]} &- \ce{[H2A-][H3O+]} = 0\\ k_{a2} &= \ce{\frac{[HA^2-][H3O+]}{[H2A-]}} \\ k_{a2} \ce{[H2A-]} &- \ce{[HA^2-][H3O+]} = 0\\ k_{a3} &= \ce{\frac{[A^3-][H3O+]}{[HA-]}} \\ k_{a3} \ce{[HA-]} &- \ce{[A^3-][H3O+]} = 0\\ \ce{[H3A] + [H2A-] &+ [HA2-] + [A3-] = [H3A]0} = c_1\\ \end{align} Which solves to: \begin{align} \ce{[H3A]} &= \frac{c_1 x^3}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\ \ce{[H2A-]} &= \frac{c_1 k_{a1} x^2}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\ \ce{[HA^2-]} &= \frac{c_1 k_{a1} k_{a2} x}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\ \ce{[A^3-]} &= \frac{c_1 k_{a1} k_{a2} k_{a3}}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \end{align}

Sodium bicarbonate, to be protonated or deprotonated: $$ \ce{[HB^-] + [H2O] <=> [B^2-] + [H3O+]\\ [HB^-] + [H2O] <=> [H2B] + [OH-]} $$ \begin{align} k_{a4} &= \ce{\frac{[B^2-][H3O+]}{[HB-]}} \\ k_{a4} \ce{[HB-]} &- \ce{[B^2-][H3O+]} = 0\\ k_{b1} &= \ce{\frac{[H2B][OH-]}{[HB-]}} \\ k_{b1} \ce{[HB-]} &- \ce{[H2B][OH-]} = 0\\ \ce{[H2B] &+ \ce{[HB-] + [B^2-] = [B]0}} = c_2\\ \end{align} Which solves to: \begin{align} \ce{[H2B]} &= \frac{c_2 x k_{b1}}{x k_{b1} + x k_W/x + k_{a4} k_W/x} = \frac{c_2 x k_{b1}}{x k_{b1} + k_W + k_{a4} k_W/x} \\ \ce{[HB-]} &= \frac{c_2 x k_W/x}{x k_{b1} + x k_W/x + k_{a4} k_W/x} = \frac{c_2 k_W}{x k_{b1} + k_W + k_{a4} k_W/x} \\ \ce{[B^2-]} &= \frac{c_2 k_{a2} k_W/x}{x k_{b1} + x k_W/x + k_{a4} k_W/x} = \frac{c_2 k_{a2} k_W/x}{x k_{b1} + k_W + k_{a4} k_W/x}\\ \end{align}

Sodium carbonate, to be doubly protonated: $$ \ce{[C] + [H2O] <=> [CH+] + [OH-]\\ [CH+] + [H2O] <=> [CH2^2+] + [OH-]} $$ \begin{align} k_{b2} &= \ce{\frac{[CH+][OH-]}{[C]}} \\ k_{b2} \ce{[C]} &- \ce{[CH+][OH-]} = 0\\ k_{b3} &= \ce{\frac{[CH2^2+][OH-]}{[CH+]}} \\ k_{b3} \ce{[CH+]} &- \ce{[CH2^2+][OH-]} = 0\\ \ce{[C] &+ [CH+] + [CH2^2+] = [C]0} = c_3\\ \end{align} Which solves to: \begin{align} \ce{[B]} &= \frac{c_3 (k_W/x)^2}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2} \\ \ce{[BH+]} &= \frac{c_3 k_{b2} k_W/x}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2} \\ \ce{[BH^2+]} &= \frac{c_3 k_{b2} k_{b3}}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2}\\ \end{align}

Putting it all together (it looks a bit different, though, as it is copied from a simplified form of Mathematica): $$x=\frac{k_W}{x} + c_1 \frac{3 k_{a1} k_{a2} k_{a3}+2 x k_{a1} k_{a2}+x^2 k_{a1}}{k_{a1} k_{a2} k_{a3}+x k_{a1} k_{a2}+x^2 k_{a1}+x^3} + c_2 \left(\frac{k_{a4} k_W}{k_{a4} k_W+x^2 k_{b1}+x k_W}-\frac{x^2 k_{b1}}{k_{a4} k_W+x^2 k_{b1}+x k_W}\right) - c_3 \frac{2 x^2 k_{b2} k_{b3}+k_W x k_{b2}}{x^2 k_{b2} k_{b3}+k_W x k_{b2}+k_w^2}$$

Solving this little monster for $x$ and calculating its negative decadic logarithm gives you the wanted pH of your system, which is 5.65.

As you mention, further thinking about the problem makes it possible to simplify it a litle bit. \begin{align} \ce{[H+] &= [OH-] + [H2A-] + 2[HA^2-] + 3[A^3-] + [B^2-] - [H2B] - [CH+] - 2 [CH2^2+]} \\ \ce{[H+] &= [OH-] + [H2A-] + 2[HA^2-] + 3[A^3-] + [CO3^2-]_B - [H2CO3]_B - [HCO3-]_C - 2 [H2CO3]_C} \\ x &= \frac{k_W}{x} + c_1 \frac{x^2 k_{a1} + 2 x k_{a1} k_{a2} + 3 k_{a1} k_{a2} k_{a3}}{x^3 + x^2 k_{a1} + x k_{a1} k_{a2} + k_{a1} k_{a2} k_{a3}} + \frac{-(c_2 + 2 c_3) x^2 - c_3 x k_{a4} + 2 c_2 k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} \\ \end{align}

This does not change the resulting pH value and thus it is not more accurate in my eyes, it only simplifies (admittedly quite a bit) the equation to be solved.

If we assume your initial pH value to be exactly 6 and that is caused by CO2, then we can add it to the equation. $$x = \frac{k_W}{x} + c_1 \frac{x^2 k_{a1} + 2 x k_{a1} k_{a2} + 3 k_{a1} k_{a2} k_{a3}}{x^3 + x^2 k_{a1} + x k_{a1} k_{a2} + k_{a1} k_{a2} k_{a3}} + \frac{-(c_2 + 2 c_3) x^2 + (c_4 - c_3) x k_{a4} + 2 (c_2 + c_4) k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} $$

Where $c_4$ can be calculated by rearranging the proton concentration equation for solely carbonic acid: \begin{align} x &= \frac{k_W}{x} + c_4 \frac{x k_{a4} + 2 k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} \\ c_4 &= \left(x - \frac{k_W}{x}\right) \left(\frac{x^2 + x k_{a4} + k_{a4} k_{a5}}{x k_{a4} + 2 k_{a4} k_{a5}}\right) \\ c_4 &= 3.21119 \cdot 10^{-6} \end{align}

Without initial pH 6: 5.6467
With initial pH 6: 5.6465

  • $\begingroup$ Thanks for the answer. There's a few points I'd like to get some clarification. 1/ How do you get that [H+] is equal to that sum? 2/ Where does the formula to get ions concentration come from? $\endgroup$
    – Gnad
    Commented Jun 26, 2018 at 12:28
  • $\begingroup$ 3/ Also, isn't it sodium carbonate that could dissociate twice (instead of sodium hydrogen carbonate)? So B(1)H+ is H2CO3, B(2)H+ is HCO3-, B(2)H2+ is H2CO3? $\endgroup$
    – Gnad
    Commented Jun 27, 2018 at 1:30
  • $\begingroup$ @Gnad yes, I mixed this up. For your other questions, I'll try to add details. $\endgroup$ Commented Jun 27, 2018 at 8:01
  • 1
    $\begingroup$ I admire your fortitude in calculating this. However you overlooked one point. The OP stated that the water from which the solution was made had a pH of 6. I'd guess from CO2 from air, but there is no way to be sure. Thus the buffer capacity at pH 6 is unknown. $\endgroup$
    – MaxW
    Commented Jun 28, 2018 at 15:09
  • 1
    $\begingroup$ @MaxW you are totally right, but this is only solvable theoretically, if it is told where this pH comes from. $\endgroup$ Commented Jun 28, 2018 at 16:21

Not really an answer but a long comment.

User ph13 answered the question with the generalized formula, then solved the equations computationally. A true tour de force.

However the particular problem could be simplified a bit using some chemical intuition. Using ph13's notation:

(1) The pKa1 of citric acid is so much greater than pKa1 of bicarbonate that essentially no fully protonated citric acid will be left. Likewise it is clear that the solution will be acidic enough that no unprotonated citric acid will be present.

$$\ce{[H3A]} \approx 0 \text{ and }\ce{[A^{3-}]} \approx 0$$


$$ \ce{[H2A^-] + [HA^{2-}]} = c_1$$

(2) There is enough citric acid so that all the carbonate will be protonated at least once. So carbonate can be neglected in the figuring total carbonate species.

$$\ce{[B^{2-}]} \approx 0$$


$$\ce{[HB^-] + [H2B]} = c_2$$


$$\ce{[C]} \approx 0$$


$$\ce{[CH^+] + [CH^{2+}]} = c_3$$

(3) Lastly separating sodium bicarbonate species concentrations ($\ce{[B^{2-}], [HB^-], [H2B]}$) and the sodium carbonate species concentrations ([C] + [CH^+] + [CH2^+]) makes no sense. In solution a carbonate ion, the bicarbonate ion and the carbonic acid can't tell from what solid they came. The gist is that the total of all carbonate species must equal 0.44 millimoles.

(4) As I pointed out in a comment on ph13's answer the 100 ml of pH 6 water is a mystery in this. Some acid is present in the original solution and it isn't clear which one, nor how much. Essentially the buffer capacity of the 100 ml of water is unknown.

(5) A pH of 5.65 is somewhat fortuitous. It is the pH at which carbon dioxide is in equilibrium between its atmospheric phase and being dissolved in water. Any more basic and the solution would draw CO2 from atmosphere. Any more acidic and the solution would effervesce CO2. Carbonic acid has an odd equilibrium in water.

(6) Method of preparation could also have an effect. If you first add citric acid to water then sodium bicarbonate the solution will effervesce $\ce{CO2}$. If you then add sodium carbonate the resulting solution will be more basic than 5.65.

Still the equations are messy and would be easiest solved numerically (trial and error or by computer).

  • $\begingroup$ 1/ Actually , according to my calculation (done from the equations provided by @pH13), [A3−] is around 2.75 mM and [H2A−] is around 1.98 mM. Now when we notice that the amount of sodium bicarbonate + carbonate is more than twice and less than 3 times of citric acid, I think we would expect the reaction to have an equilibrium between [HA2-] and [A3−]. 3/ Agreed. That's what I was trying to say to @pH13. 4+5/ So let's assume that pH 6 of initial water is CO2 absorbed from atmosphere. How does that extra acidic affect the calculation? $\endgroup$
    – Gnad
    Commented Jun 29, 2018 at 0:55
  • $\begingroup$ Thanks for the correction about $\ce{[A^{3-}]}$. Nice plot here en.wikipedia.org/wiki/File:Citric_acid_speciation.png // ph13's calculation is as good as can be done with the information given. $\endgroup$
    – MaxW
    Commented Jun 29, 2018 at 1:16

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