# Van't Hoff factor for citric and phosphoric acids

I am trying to determine the van't Hoff Factor for citric and phosphoric acids.

When phosphoric acid is in aqueous solution, it dissociates as follows:

$$\ce{H3PO4 ->H2PO4- + H3O+}$$

Does this mean its van't Hoff factor is 2? For citric acid I cannot find what it dissociates into anywhere, so I am unable to determine the van't Hoff factor.

• No. Phosphoric acid does not technically meet the arbitrary definition of a strong acid, and the dissociation is not as high as for something like hydrochloric acid. The van't Hoff factor will still be fairly high, but not 2. Ionization of the other species will be relatively low though they in principle will have an effect on the van't Hoff factor. The same idea applies to citric acid. – Zhe Jun 7 '17 at 15:31
• Thank you, but is there as way I can get the actual van't hoff factor or at least a number that is close to it? – user46204 Jun 7 '17 at 15:44
• You can compute from the pKa. – Zhe Jun 7 '17 at 16:30

## 1 Answer

The Van't Hoff factor isn't an intrinsic property of any species - it depends on the dissociation degree of an electrolyte. Both of those acids are weak and triprotic, meaning they can lose up to three protons in solution (depending on the pH). Therefore, the Van't Hoff factor will likewise change depending on the pH.

1) If the acid is the only species in solution, and since it's a weak acid, you need to use the same formula as for any weak electrolyte:

$i=1+\alpha$

You can figure out the dissociation factor, $\alpha$, from the dissociation constant:

$K_1=\cfrac{[H_2A^-][H^+]}{[H_3A]}=\cfrac{(\alpha C_{H_3A})^2}{(1-\alpha)C_{H_3A}}=\cfrac{\alpha^2}{1-\alpha}C_{H_3A}$

where $C_{H_3A}$ is the formal concentration of the acid.

2) If you have more species in solution, it will depend on the pH of the solution. In general, you'll have this expression:

$i=\cfrac{[H_3A]+[H_2A^-]+[HA^{2-}]+[A^{3-}]+[H^+]}{C_{H_3A}}$

After some tedious substitutions you'll get

$i=1+\cfrac{K_1}{[H^+]+K_1}+\cfrac{K_2}{[H^+]+K_2}+\cfrac{K_3}{[H^+]+K_3}$

Some relevant values:

$pH=0 \rightarrow i \approx 1$

$pH=pK_1 \rightarrow i = 1.5$

$pH=pK_2 \rightarrow i = 2.5$

$pH=pK_3 \rightarrow i = 3.5$

$pH=14 \rightarrow i \approx 4$

and it changes logarithmically in between.