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

Feb 16, 2018

$\text{pH} = 3.577$

#### Explanation:

Potassium hydrogen phthalate, or simply $\text{KHP}$, is actually an acidic salt because its anion, the hydrogen phthalate anion, ${\text{HP}}^{-}$, 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.

${\text{KHP"_ ((aq)) -> "K"_ ((aq))^(+) + "HP}}_{\left(a q\right)}^{-}$

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

${\text{HP"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "P"_ ((aq))^(2-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

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

${K}_{a} = {10}^{- \text{p} {K}_{a}}$

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 $x$ $\text{M}$ of hydrogen phthalate anion ionize, you will get $x$ "M" of phthalate anions and $x$ $\text{M}$ of hydronium cations.

Moreover, the concentration of the hydrogen phthalate anions will decrease by $x$ $\text{M}$, 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

${K}_{a} = \left(\left[{\text{P"^(2-)] * ["H"_3"O"^(+)])/(["HP}}^{-}\right]\right)$

to get

${10}^{- \text{p} {K}_{a}} = \frac{x \cdot x}{0.01763 - x}$

${10}^{- \text{p} {K}_{a}} = {x}^{2} / \left(0.01763 - x\right)$

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

$0.01763 - x \approx 0.01763$

This means that you have

${10}^{- \text{p} {K}_{a}} = {x}^{2} / 0.01763$

which will get you

$x = \sqrt{0.01763 \cdot {10}^{- 5.4}} = 0.00026493$

Since $x$ $\text{M}$ represents the equilibrium concentration of hydronium cations, you can say that you have

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

As you know, the $\text{pH}$ of the solution can be calculated using the equation

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

In your case, the $\text{pH}$ of the solution will be

$\text{pH} = - \log \left(0.00026493\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{3.577}}}$

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.