Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# What is the equilibrium constant for the weak acid KHP?

Potassium hydrogen phthalate ("KHP") has the "C"_8"H"_5"KO"_4 and is known to be a .

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

KHP_((aq)) rightleftharpoons K_((aq))^(+) + HP_((aq))^(-), followed by

HP_((aq))^(-) + H_2O_((l)) rightleftharpoons H_3^(+)O_((aq)) + P_((aq))^(2-)

The expression for the reaction's is

K_(eq) = ([H_3^(+)O] * [P^(2-)])/([HP^(-)] * [H_2O])

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, "K"_"a".

K_a = ([H_3^(+)O] * [P^(2-)])/([HP^(-)])

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

K_a = 3.9 * 10^(-6), which confirms that "KHP" is a weak acid.

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