# Question 15adf

Apr 8, 2016

$\text{pH} = 7.10$

#### Explanation:

You're dealing with a buffer solution that contains dihydrogen phosphate, ${\text{H"_2"PO}}_{4}^{-}$, which will act as a weak acid, and hydrogen phosphate, ${\text{HPO}}_{4}^{2 -}$, which is its conjugate base.

As you can see, the weak acid will be delivered to the solution by monopotassium phosphate, ${\text{KH"_2"PO}}_{4}$, and the conjugate base will be delivered to the solution by dipotassium phosphate, ${\text{K"_2"HPO}}_{4}$, both soluble salts that dissociate completely in aqueous solution.

The following equilibrium will be established in solution

${\text{H"_ 2"PO"_ (4(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HPO"_ (4(aq))^(2-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

Before moving on, calculate the concentrations of the weak acid and conjugate base by using the fact that both monopotassium phosphate and dipotassium phosphate dissociate in a $1 : 1$ mole ratio to form ${\text{H"_2"PO}}_{4}^{-}$ and ${\text{HPO}}_{4}^{2 -}$, respectively.

Start by using the molar masses of the two salts to figure out how many moles of each you get in those $\text{15.00-g}$ samples

15.00 color(red)(cancel(color(black)("g"))) * ("1 mole KH"_2"PO"_4)/(136.086color(red)(cancel(color(black)("g")))) = "0.11022 moles KH"_2"PO"_4

15.00 color(red)(cancel(color(black)("g"))) * ("1 mole K"_2"HPO"_4)/(174.2color(red)(cancel(color(black)("g")))) = "0.086108 moles K"_2"HPO"_4

Now, because the total volume of the solution is said to be equal to ${\text{1.00 dm}}^{3}$, which is equivalent to $\text{1.00 L}$, you can treat number of moles and molarity interchangeably.

As you know the molarity of a solution tells you how many moles of solute you get per cubic decimeter of solution. In this case, you know that ${\text{1.00 dm}}^{3}$ of solution will contain

${n}_{{H}_{2} P {O}_{4}^{-}} = {n}_{K {H}_{2} P {O}_{4}} = {\text{0.11022 moles H"_2"PO}}_{4}^{-}$

${n}_{H P {O}_{4}^{2 -}} = {n}_{{K}_{2} H P {O}_{4}} = {\text{0.086108 moles HPO}}_{4}^{2 -}$

which implies that you have

["H"_2"PO"_4^(-)] = "0.11022 mol dm"^(-3)

["HPO"_4^(2-)] = "0.086108 mol dm"^(-3)

The pH of a buffer solution that contains a weak acid and its conjugate base can be calculated using the Henderson - Hasselbalch equation, which looks like this

color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))

Notice that when you have equal concentrations of weak acid and conjugate base, the pH of the buffer is equal to the $p {K}_{a}$ of the acid.

In your case, the buffer contains more weak acid than conjugate, so right from the start you can predict that its pH will be lower than the $P {K}_{a}$ of the acid.

In your case, the H - H equation will look like this

"pH" = 7.21 + log( (["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)]))

Plug in your values to get

"pH" = 7.21 + log( (0.086108 color(red)(cancel(color(black)("mol dm"^(-3)))))/(0.11022color(red)(cancel(color(black)("mol dm"^(-3)))))) = color(green)(|bar(ul(color(white)(a/a)7.10color(white)(a/a)|)))#

As predicted, the buffer has a pH that's lower than the $p {K}_{a}$ of the acid.