Question

How to Calculate the pH of 0.180 g of potassium biphthalate (pKa = 5.4) in 50.0 mL of water?

Answer 1

`"pH" = 3.577`

Explanation:

Potassium hydrogen phthalate, or simply `"KHP"`, is actually an acidic salt because its anion, the hydrogen phthalate anion, `"HP"^(-)`, acts as a weak acid in aqueous solution.

So when you dissolve the salt in water, you get potassium cations and hydrogen phthalate anions in `1:1` mole ratios.

`"KHP"_ ((aq)) -> "K"_ ((aq))^(+) + "HP"_ ((aq))^(-)`

Now, the hydrogen phthalate will partially ionize to produce phthalate anions and hydronium cations.

`"HP"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "P"_ ((aq))^(2-) + "H"_ 3"O"_ ((aq))^(+)`

By definition, the acid dissociation constant for the hydrogen phthalate anion is given by

`K_a = 10^(-"p"K_a)`

Use the molar mass of potassium hydrogen phthalate to calculate the number of moles present in the sample

`0.180 color(red)(cancel(color(black)("g"))) * "1 mole KHP"/(204.22 color(red)(cancel(color(black)("g")))) = "0.0008814 moles KHP"`

This means that the solution will contain `0.0008814` moles of hydrogen phthalate anions, which would have an initial concentration of--you can assume that the volume of the solution will be equal to the volume of water.

`["HP"^(-)] = "0.0008814 moles"/(50.0 * 10^(-3) quad "L") = "0.01763 M"`

Now, if `x` `"M"` of hydrogen phthalate anion ionize, you will get `x` `"M`" of phthalate anions and `x` `"M"` of hydronium cations.

Moreover, the concentration of the hydrogen phthalate anions will decrease by `x` `"M"`, so you can say that, at equilibrium, the solution will contain

`["P"^(2-)] = ["H"_3"O"^(+)] = x quad "M"`

`["HP"^(-)] = (0.01763 - x) quad "M"`

Plug this into the expression of the acid dissociation constant

`K_a = (["P"^(2-)] * ["H"_3"O"^(+)])/(["HP"^(-)])`

to get

`10^(-"p"K_a) = (x * x)/(0.01763 - x)`

`10^(-"p"K_a) = x^2/(0.01763 - x)`

Now, because the value of the acid dissociation constant is significantly smaller than the initial concentration of the hydrogen phthalate anions, you can use the approximation

`0.01763 - x ~~ 0.01763`

This means that you have

`10^(-"p"K_a) = x^2/0.01763`

which will get you

`x = sqrt(0.01763 * 10^(-5.4)) = 0.00026493`

Since `x` `"M"` represents the equilibrium concentration of hydronium cations, you can say that you have

`["H"_3"O"^(+)] = "0.00026493 M"`

As you know, the `"pH"` of the solution can be calculated using the equation

`color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))`

In your case, the `"pH"` of the solution will be

`"pH" = - log(0.00026493) = color(darkgreen)(ul(color(black)(3.577)))`

The answer is rounded to three decimal places, the number of sig figs you have for the mass of potassium hydrogen phthalate and the volume of the solution.