Phosphoric acid is a triprotic acid with the following pKa values:

pka1: 2.148 pka2: 7.198 pka3: 12.375

You wish to prepare 1.000 L of a 0.0100 M phosphate buffer at pH 7.45. To do this, you choose to use mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?

?? grams NaH2PO4

?? grams Na2HPO4

Hint: Use the Henderson-Hasselbalch equation to get the molar ratio of Na2HPO4 to NaH2PO4 required, then the fraction of each form from the ratio. The total moles needed will be 1.000 L × 0.0100 M = 0.0100 moles. Use the formula mass to calculate the mass needed. (FM NaH2PO4 = 119.98; FM Na2HPO4 = 141.96).

I am completely lost on how to solve this problem.

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asked by Sam

Feb 22, 2013

pH=pka+log({A-]/[HA])

pH=7.45

pka=12.375

Solve for ratio:

10^(7.45-12.375)={A-]/[HA]=1.19×10^-5

Since you have a total of 0.01 moles,

1.19×10^-5*(0.01moles)= moles of Na2HPO4

0.01 moles-moles of Na2HPO4= moles of NaH2PO4

Use the formula weights to solve for the number for each that Dr. Bob222 gave you.

**** Not sure about that pKa value that I chose.

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posted by Devron

Feb 22, 2013

The correct pKa value to choose is pK2.

Use the HH equation to solve for the ratio base/acid.

One equation you need is base/acid.

The other equation you need is

base + acid = 0.01

That two equations in two unknowns; solve for acid concn and base concn and convert to grams. Post your work if you gets stuck.

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posted by DrBob222

Feb 22, 2013

Setup is correct, but substitute 7.198. That should change 1.19×10^-5 to 1.78 and do what Dr.Bob222 told you to do.

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posted by Devron

Feb 22, 2013

if you use the pk2 wouldnt you get a negative value for nah2po4?

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posted by scott

Oct 27, 2014

After you get the ratio. From the

0.01 moles-moles of Na2HPO4= moles of NaH2PO4

=> (0.01 – Na2HPO4)/(Na2HPO4) = (ratio of pka2 from hasselbalch)

unit will be mol, convert to g

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posted by Shiro

Oct 26, 2016