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

##### 1 Answer

#### Answer:

#### Explanation:

Potassium hydrogen phthalate, or simply **weak acid** in aqueous solution.

So when you dissolve the salt in water, you get potassium cations and hydrogen phthalate anions in **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 **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

Moreover, the concentration of the hydrogen phthalate anions will **decrease** by

#["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

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

As you know, the

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

In your case, the

#"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.