# Citric acid

1. Feb 25, 2008

### Engbryg

How can I calculate the amount of citric acid to add to a solution of a given pH, to achieve another value of pH?

2. Feb 25, 2008

### symbolipoint

Experimentally, use small increments of citric acid and monitor pH; then scale the ratio upward to the size of solution you want to adjust.

Theoretically, use polyprotic weak acid equilibrium dissociation constants...
... loosing skills here, so hope you want experimental general answer.

3. Feb 25, 2008

### Engbryg

No, I was hoping for a ready-to-use formular, i.e. x mg of citric acid to decrease from pH 8 to pH 5.2.

4. Feb 25, 2008

### symbolipoint

You best probably cannot find such a formula to be too simple. You may need to know the source of the alkalinity. If the source of alkalinity is just sodium hydroxide, then maybe you could rely on dissociation constant information with equilibrium calculations to derive a formula. Maybe someone who is currently more skilled in this area can comment further, since my proficiency is no longer strong in this area.

Actually, you could even try measured additions of citric acid of known concentration from a buret while monitoring pH; you can then pre-design your proportion formula.

5. Feb 26, 2008

### AbedeuS

To be honest, I just threw this out, im not even 100% sure if any of this maths is right as it seemed to be a slippery slope trying to solve this one.

I just put this here to show that the community here do try and make an effort, but this one is whey out of my range.

pH is:
$$pH=-log([H^+])$$

Therefore:
$$[H^+] = 10^{-pH}$$
$$\Delta [H^+]=10^{-pH_2}-10^{-pH_1}$$

Intermediate example you chosen: pH 8 -> pH 5.2
$$[H^+_1] = 10^{-8}$$
$$[H^+_2] = 10^{-5.2}$$

So if we have a solution that is at pH8 we need to increase the $$H^+$$ concentration by $$10^{-5.2}-10^{-8}$$

Citric acid isnt a strong acid, so its Ka is < 1. We're aiming for the above mentioned concentration change so:

$$K_a(citric) = \frac{[X-][H^+]}{[Citric acid added]-[X^-]}$$

Rearranging, remembering that [X-] = [H+] if no other acids/bases are present otherwise youll get a buffer case concentration wise:
$$[\Delta H^+]^2+[\Delta H^+]K_a = [Citric acid added]K_a$$

You must remember the volume of liquid already present might not be equal to a $$dm^3$$ so:

$$[Citric acid added] = \frac{n}{v} = \frac{m}{192.123v}$$

Volume of solution in $$dm^3$$
Mass in grams

So inserting all what we know:
$$[\Delta H^+]^2+[\Delta H^+]K_a = \frac{mK_a}{192.123V}$$

Rearranging it for m to be the subject

$$\frac{192.123V}{K_a}([\Delta H^+]^2 + [\Delta H^+]K_a) = m$$

Since we found $$[\Delta H^+]$$ before we can sub that in:

$$\frac{192.123V}{K_a}((e^{-2*pH_2}-e^{-2*pH_1})+(e^{-pH_2}-e^{-pH_1})K_a) = m$$

As this is a quadratic equation we make $$[\Delta H^+]$$ the subject as such (if your trying to find the pH change on addition of m amount of acid to v):

$$[\Delta H^+] = \frac{-K_a+\sqrt{K_a^2 + 4*\frac{mK_a}{192.123V}}}{2}$$

If you want to know the pH change from adding mass m to volume V then:

$$\Delta pH = log_10 \left[ \frac{-K_a+\sqrt{K_a^2+4*\frac{mK_a}{192.123V}}}{2}\right]$$

The left side of the equation is decided by you, then the right side is found out, where:
$$K_a$$ Ka of the acid in question
$$m$$ mass of solid acid added that is changing the pH
$$V$$ Volume of solvent/solution that your adding the solid acid too

To be honest, I just threw this out, im not even 100% sure if any of this maths is right as it seemed to be a slippery slope trying to solve this one.

This not only goes off the assumption of a monoprotic acid, but a monoprotic acid that is not in a salt solution (buffer possibilities), as you can see i messed my head up trying to calculate this extremely "easy" example (since I thought I'd do this in no time whatsoever) imagine trying to do this under more realistic conditions.

Last edited: Feb 26, 2008
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