You wish to make a 1 liter solution of 0.1 M sodium phosphate buffer (pH 6.80).

Question

You wish to make a 1 liter solution of 0.1 M sodium phosphate buffer (pH 6.80). Phosphoric acid (H3PO4) has three protons that may dissociate, with pKa values of 2.12, 7.21, and 12.32 Calculate the total number of moles of NaOH that would have to be added to yield a 6.80 pH value. Select one: O a. The total amount of NaOH would be 0.072 moles O b. The amount of NaOH required is very difficult to calculate, because it is necessary to know both the concentration of phosphoric acid and sodium phosphate present, and therefore there are more variables than equations. O c. The total amount of NaOH would be 0.028 moles. O d. The total amount of NaOH would be 0.172 moles O e. The total amount of NaOH would be 0.128 moles

Explanation / Answer

Firstly you need to do the calculations via Handerson Hasselbach equation:

pKa's of phosphoric acid are 2.3, 7.21 and 12.35. If required pH is 6.8, then, 7.21 will be used. This means, monosodium dihydrogen phosphate and disodiumium monohydrogen phosphate (diprotic (H2PO4-) and monoprotic (HPO42-) potassium salts) will be used.

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

Now, for A- you put Na2HPO4 concentration, and for HA you put NaH2PO4 concentration:

6.8= 7.21 + log(A/HA)

log(A/HA)=-0.41

A/HA = 0.389

So, now you know the fold difference between these salts, and the Molarity of the solution would be given to you, and you can calculate the required amount:

HA+A=0.1

A/HA=0.389=>A=0.389HA

HA+A=0.1

0.389HA+HA=0.1

=>HA=0.1/1.389=0.072M

A=0.1-.072=0.028M

This means you need to put 0.072 moles of NaH2PO4 and 0.028 moles of Na2HPO4 salts into 1 liters of solution.

0.072 moles of NaH2PO4 Contains Na+ =0.072mole

0.028 moles of Na2HPO4  Contains Na+=2 x 0.028=0.056mole

no. of the mole of Na+= no. of the mole of OH-

so total no. of the mole of NaOH required=0.072+0.056=0.128moles

option e is the right answer.

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