How would you prepare 1 L of a 0.05 m Phosphate buffer at ph 7.5 using crystalli

Question

How would you prepare 1 L of a 0.05 m Phosphate buffer at ph 7.5 using crystalline K2HPO4 and a solution of 1 M HCL?

So far I have gotten that I need to add 8.71g of K2PO4- . Because I want a pH of 7.5, but the closest acid I can use has a pH of 7.2 (according to my table):

7.5=7.2+log n(HPO42-)/n(H2PO4-)...some simplifying...

0.3=log  n(HPO42-)/n(H2PO4-)

2.00=n(HPO42-)/n(H2PO4-)

And this next part has me confused. My Proff.'s notes says that we have 2 equations

n(HPO42-)= 2 x n(H2PO4-) and n(HPO42-)+n(HPO4-)= 0.050 and to set them equal to get

3n(H2PO4-)=0.050 mol, giving that the volume needed is 1.67x10-2L. My question comes when solving the two equations, as I donnot get the same answer he has. Maybe my algebra is a little rust, but I just can't over come that part of this problem. How can you solve the two equations? Thank you for your help!

Explanation / Answer

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 HPO42-. The relevant pKa is 7.2 and the pH =7.5.

To get started, let:

x=[HPO42-]/[H2PO4-]

Then:

7.5=7.2+log(x)

0.3=log(x)

100.3=x

2=x=[HPO42-]/[H2PO4-] (Eqn. 1)

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

[HPO42-] +[H2PO4-] = 0.05 M

[H2PO4-] = 0.05M -[HPO42-] (Eqn. 2)

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

2=[HPO42-]/{0.05M-[HPO42-]}

0.1=3[HPO42-]

[HPO42-]=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 HPO42- = (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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