# How does pH of soda drink vary with time

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.

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.

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:

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

where

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

Then

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

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

• 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. – Aurora Borealis Feb 22 '20 at 15:33
• See the answer addition. – Poutnik Feb 22 '20 at 15:48
• Thank you, additionally i do not understand how equation (2) became (3), where did 2.303 come from? – Aurora Borealis Feb 29 '20 at 11:06
• 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. – Poutnik Feb 29 '20 at 11:25
• Oh yea thats true, my bad – Aurora Borealis Feb 29 '20 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.