Question

Question #15adf

Answer 1

`"pH" = 7.10`

Explanation:

You're dealing with a buffer solution that contains dihydrogen phosphate, `"H"_2"PO"_4^(-)`, which will act as a weak acid, and hydrogen phosphate, `"HPO"_4^(2-)`, which is its conjugate base.

As you can see, the weak acid will be delivered to the solution by monopotassium phosphate, `"KH"_2"PO"_4`, and the conjugate base will be delivered to the solution by dipotassium phosphate, `"K"_2"HPO"_4`, both soluble salts that dissociate completely in aqueous solution.

The following equilibrium will be established in solution

`"H"_ 2"PO"_ (4(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HPO"_ (4(aq))^(2-) + "H"_ 3"O"_ ((aq))^(+)`

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 `"H"_2"PO"_4^(-)` and `"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 `"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 `"1.00 dm"^3`, which is equivalent to `"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 `"1.00 dm"^3` of solution will contain

`n_(H_2PO_4^(-)) = n_(KH_2PO_4) = "0.11022 moles H"_2"PO"_4^(-)`

`n_(HPO_4^(2-)) = n_(K_2HPO_4) = "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 `pK_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 `PK_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 `pK_a` of the acid.