Basically I have done an experiment where I measure the pH of my soda drink at a specific temperature over $\pu{45 min}$. At every temperature, the $\mathrm{pH}$ of my soda drink increases, but much faster for higher temperatures.

For instance, I measured at $\pu{25 ^\circ C}$ how fast the pH increases of my soda drink over $\pu{45 min}$. Next I measured at $\pu{35 ^\circ C}$ how fast the $\mathrm{pH}$ increases of my soda drink over $\pu{45 min}$. Continued for the next 4 temperatures.

I plotted the graph of how $\mathrm{pH}$ increases with time for 6 temperatures (getting me 6 graphs), but I have difficulty finding out the relationship I have is linear or exponential. The current trend I have only seems to be linear from the $\pu{45 min}$ time period I performed my experiment for, but I was wondering, if it just looks linear because my experiment was faulty, or it is actually supposed to be linear in relationship. I basically do not know if there is a theory which states that the relationship I am looking for is or is not linear.

I will appreciate the assistance.


2 Answers 2


I will make few simplifications:

  • I will neglect presence of other acids like citric or phosphoric acid, considering just water and carbon dioxide.

  • I will use the simplified exquation of $\mathrm{pH}$ of a weak acid ( :

$$\mathrm{pH}=\frac 12 (\mathrm{p}K_\mathrm{a} - \log {[\ce{CO2}]}) \tag{1}$$

  • I will consider $\ce{CO2}$ escaping as the exponential process of the 1st order kinetics.

$$[\ce{CO2}]=A + B \cdot \exp{(-Ct)} \tag{1a}$$

  • I will neglect the final equilibrium $\ce{CO2}$ concentration, compared to initial one:

$$[\ce{CO2}]=B \cdot \exp{(-Ct)} \tag{1b}$$

We can realize that the exponential function for $\ce{CO2}$ escaping kind of nullifies the logarithm function of $\mathrm{pH}$ definition:

$$\mathrm{pH}=\frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {\left(B \cdot \exp{(-Ct)}\right)}\right) \tag{2}$$

$$\mathrm{pH} \simeq \frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {B} + \frac1{2.303}Ct\right) \tag{3}$$

Therefore, $\mathrm{pH}=f(t)$ is approximately linear.

It is not linear if we consider all factors, especially if other acids are present.

Additional requested clarification:

The equation (1) is quite notoriously known simplified equation for $\mathrm{pH}$ of weak acid. It can be directly derived from definition of a dissociation constant of an acid:

$$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}} \tag{4}$$

involving 2 simplifications:

  • solution is acidic enough to ignore water auto-dissociation:

$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{\ce{[HA]}} \tag{5}$$

  • solution contains enough of a weak acid, so it's majority is not dissociated:

$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{c} \tag{6}$$


$$c=\ce{[HA]}+\ce{[A-]} \tag{7}$$


$$\ce{[H+]}=\sqrt{ K_\mathrm{a} \cdot c} \tag{8}$$

After logaritmization of (8), we get (1).

  • $\begingroup$ May I know how that equation you gave regarding the 1/2 (expression) came about in the very first line of the post? I would appreciate it. $\endgroup$ Commented Feb 22, 2020 at 15:33
  • 1
    $\begingroup$ See the answer addition. $\endgroup$
    – Poutnik
    Commented Feb 22, 2020 at 15:48
  • $\begingroup$ Thank you, additionally i do not understand how equation (2) became (3), where did 2.303 come from? $\endgroup$ Commented Feb 29, 2020 at 11:06
  • $\begingroup$ As an undergraduate majoring in mathematics, you should know that in chemistry context, log means log10, log(exp(x))= ln(exp(x))/ln(10)~=x/2.303. Additionally, the omission of A relates to omission of the final CO2 equilibrium content. $\endgroup$
    – Poutnik
    Commented Feb 29, 2020 at 11:25
  • $\begingroup$ Oh yea thats true, my bad $\endgroup$ Commented Feb 29, 2020 at 16:14

Pure water has a pH that goes form 7.00 at 25°C, to 6.92 at 30°C and to 6.13 at 100°C. As you obtain an effect which is contrary to this one, it means that your solution is loosing dissolved $\ce{CO2}$, so that the solution becomes more and more alcaline (or basic). After some time, you may obtain a precipitate of $\ce{CaCO3}$, depending on the composition of your soda.


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