Solution to PS 1 #11

To solve this problem first use the Henderson-Hassalbalch equation to get the ratio of the acidic and basic forms. In this case the acidic form (HA) is H₂PO₄ ^-and the basic form (A^-) is HPO₄^2-. The relevant pKa is 7.2 and the pH =7.5.

To get started, let:

x=[HPO₄^2-]/[H₂PO₄^-]

Then:

7.5=7.2+log(x)

0.3=log(x)

10^0.3=x

2=x=[HPO₄^2-]/[H₂PO₄^-] (Eqn. 1)

We know that the total concentration of phosphate buffer is 0.05M, so

[HPO₄^2-] +[H₂PO₄^-] = 0.05 M

[H₂PO₄^-] = 0.05M -[HPO₄^2-] (Eqn. 2)

Substituting into Eqn. 2 into Eqn. 1, we get:

2=[HPO₄^2-]/{0.05M-[HPO₄^2-]}

0.1=3[HPO₄^2-]

[HPO₄^2-]=0.033 M

and [H₂PO₄^-]= 0.05 M-0.0333 M= 0.0167 M

Now convert to moles to continue:

Moles of H₂PO₄^- = (0.0167M)(1 L)= 0.0167 moles

Moles of HPO₄^2- = (0.0333M)(1L)= 0.0333 moles.

Now if we start with K₂HPO₄ and add HCl (H+) we will be converting HPO₄ into H₂PO₄. This means we have to add as many moles of HCl as we have H₂PO₄^- at pH 7.5, i.e. we have to add 0.0167 moles of HCl. Using the definition of molarity if we start with 1 M HCl we will have to add 0.0167 moles/ 1 M= 0.167 L = 16.7 mls of 1 M HCl.

Since we want the total concetration of phosphate buffer to be 0.05 M and the total volume to be 1 Liter, we need to start with 0.05 moles of K₂HPO₄. Since the molecular weight is 174 g/mole, this means we need to start with

(0.05 moles)(174 g/mole)=8.7 g of K₂HPO₄.

Answer: 8.7 g of K₂PO₄ dissolved in approximately 800 mls water, then add the 16.7 mls 1M HCl, finally bring total volume to 1 Liter with additional water.

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