How to Calculate the pH of 0.180 g of potassium biphthalate (pKa = 5.4) in 50.0 mL of water?
1 Answer
Explanation:
Potassium hydrogen phthalate, or simply
So when you dissolve the salt in water, you get potassium cations and hydrogen phthalate anions in
#"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
#["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.