What is the equilibrium constant for the weak acid KHP?

Jan 14, 2015

Potassium hydrogen phthalate ($\text{KHP}$) has the molecular formula ${\text{C"_8"H"_5"KO}}_{4}$ and is known to be a weak acid.

When placed in water, $\text{KHP}$ dissociates completely into the potassium cation ${\text{K}}^{+}$ and the hydrogen phthalate anion, "HP^(-). After the dissociation takes place, ${\text{HP}}^{-}$ reacts with water to give the hydronium cation, $\text{H"_3^(+)"O}$, and the phthalate anion, ${\text{P}}^{2 -}$. So,

$K H {P}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {K}_{\left(a q\right)}^{+} + H {P}_{\left(a q\right)}^{-}$, followed by

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

The expression for the reaction's equilibrium constant is

${K}_{e q} = \frac{\left[{H}_{3}^{+} O\right] \cdot \left[{P}^{2 -}\right]}{\left[H {P}^{-}\right] \cdot \left[{H}_{2} O\right]}$

Because the concentration of water is presumed constant and therefore not included in the expression, the equilibrium constant for this reaction is called acid dissociation constant, $\text{K"_"a}$.

${K}_{a} = \frac{\left[{H}_{3}^{+} O\right] \cdot \left[{P}^{2 -}\right]}{\left[H {P}^{-}\right]}$

Usually, this acid dissociation constant for a particular reaction will be given to you; in $\text{KHP}$'s case $\text{K"_"a}$ is equal to

${K}_{a} = 3.9 \cdot {10}^{- 6}$, which confirms that $\text{KHP}$ is a weak acid.

GIven the above expression, you can solve for $\text{K"_"a}$ by using the equilibrium concentrations of the species involved.