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### Topic: Urgent help needed - henderson hasselbalch buffers  (Read 3794 times)

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#### Bigboy0099

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##### Urgent help needed - henderson hasselbalch buffers
« on: September 09, 2017, 06:53:45 PM »
I Do not understand how cengage got "[HPO42-] = 0.0334 M and [H2PO4-] = 0.0666 " from a 0.50 mol ratio, if someone can explain the math they used it would help.

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How would you prepare 1 L of a 0.100 M phosphate buffer at pH 6.90 using crystalline Na2HPO4 and a solution of 1.0 M HCl?

In the answer boxes below, indicate how much of each species you would add to make up the solution. Some physical constants for Na2HPO4 are given in the "Hint(s)" link.
Mix together  grams of Na2HPO4,  mL of 1.0 M HCl, and  L of water.

First, the amount of Na2HPO4 to add must be calculated. To make a final concentration of 0.100 of Na2HPO4 in a final volume of 1 L of buffer, 0.100 mol of Na2HPO4 must be added. The number of grams of Na2HPO4 that the 0.100 mol represents equals:
(molar mass)×(number of moles) = (142 g/mol)×(0.100 mol) = 14.2 g

Next, note that this 0.100 M solution of Na2HPO4 will have very nearly all of its phosphate in the form of HPO42- (Na2HPO4 will dissociate completely into cations and anions), and thus will be quite basic. It is necessary to add acid to protonate some of this HPO42- to H2PO4-. To determine the amount to add, the Henderson-Hasselbalch equation can be used. This equation says that:
pH = pKa + log([HPO42-]/[H2PO4-])

Rearranging this equation gives us:
log([HPO42-]/[H2PO4-]) = pH - pKa

For the above values,
log([HPO42-]/[H2PO4-]) = 6.90 - 7.20 = -0.30

By taking the antilog of both sides of this equation, it is straightforward to calculate that:
[HPO42-]/[H2PO4-] = 10-0.30 = 0.50

The total amount of all phosphate species in the buffer must equal the amount of HPO42- that was added originally as Na2HPO4. Thus,
[HPO42-] + [H2PO4-] = 0.100 M

This equation plus the equation for the ratio [HPO42-]/[H2PO4-] gives us two equations with two unknowns, which can be solved algebraically. Solving these two equations simultaneously gives us:
[HPO42-] = 0.0334 M

and

[H2PO4-] = 0.0666 M

Given that the solution volume is 1 L, this result says that there must be 0.0334 mol of HPO42- and 0.0666 mol of H2PO4- present at pH 6.90.

To arrive at the proper ratio for [HPO42-]/[H2PO4-], the original 0.100 mol of HPO42- must be reduced to 0.0334 mol and the amount of H2PO4- must be increased to 0.0666 mol. This can be done by titrating with 0.0666 mol of HCl. This much HCl is in 67 mL of 1.0 M HCl.

The final step is to bring the buffer volume up to 1.0 L. If the assumption is made that the crystalline Na2HPO4, when it dissolves, does not measurably alter the final volume of the solution, it would be necessary to add 0.93 L of water to bring the volume to 1.0 L.

#### Borek

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##### Re: Urgent help needed - henderson hasselbalch buffers
« Reply #1 on: September 10, 2017, 03:43:02 AM »
You have two equations:

x+y=0.1
x/y=0.5

It is a simple algebra.

Plenty of resources on th web, see for example http://www.math-mate.com/chapter14_2.shtml
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info