# What is the equilibrium constant of citric acid?

Jan 14, 2015

Citric acid falls into the category of polyprotic acids, which are acids that have more than one acidic hydrogen that can react with water to produce the hydronium ion, $\text{H"_3^(+)"O}$.

Citric acid's molecular formula is ${\text{C"_6"H"_8"O}}_{7}$, and it's known as a weak organic acid. CItric acid is actually a triprotic acid, which means it has 3 acidic hydrogen atoms in its structure, as you can see below:

When placed in water, citric acid will ionize in a step-wise manner

${C}_{6} {H}_{8} {O}_{7 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {C}_{6} {H}_{7} {O}_{7 \left(a q\right)}^{-} + {H}_{3}^{+} {O}_{\left(a q\right)}$ (1)

${C}_{6} {H}_{7} {O}_{7 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {C}_{6} {H}_{6} {O}_{7 \left(a q\right)}^{2 -} + {H}_{3}^{+} {O}_{\left(a q\right)}$ (2)

${C}_{6} {H}_{6} {O}_{7 \left(a q\right)}^{2 -} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {C}_{6} {H}_{5} {O}_{7 \left(a q\right)}^{3 -} + {H}_{3}^{+} {O}_{\left(a q\right)}$ (3)

For each of these three steps we have a different value for the acid's dissociation constant, $\text{K"_"a}$. Thus,

Step (1): $\text{K"_"a1} = 7.5 \cdot {10}^{- 4}$

Step (2): $\text{K"_"a2} = 1.7 \cdot {10}^{- 5}$

Step (3): $\text{K"_"a3} = 4.0 \cdot {10}^{- 7}$

Notice that all three dissociation constants are smaller than 1, which is characteristic of a weak acid. Another interesting observation is that the dissociation constant for step (3) is very, very small, which means that the number of acid molecules that undergo ionization in this stage is, for all intended purposes, zero.