Question #15adf

1 Answer
Apr 8, 2016

Answer:

#"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.