The question:

a)Write a complete procedure for the preparation of a phosphate buffer with a pH of 7, total phosphate concentration of 0.05 M, and a total volume of 1.0 L. Your available reagents are: phosphoric acid (14.8 M) and anhydrous sodium phosphate (Na3 PO4 , MW = 164). The three pKa values for phosphoric acid are 2.15, 7.20, 12.35.

b)Calculate the pH of your buffer upon the addition of 1.0 mL of concentrated hydrobromic acid (8.89 M, pKa -8)

The way I originally solved part (a) was wrong:

using Henderson-Hasselbalch equation: pH=pka + log [A-]/[HA]

7=7.2 + log [A-]/[HA]

[A-]/[HA] = 0.6310

0.6310x+ 1x = 0.05M

x=0.0307M = [HA]

[A-]= 0.01934 M

c1v1=c2v2

(14.8M)(v1)=(0.0307M)(1.0L)

v1= 0.00207 L

0.01934 mol/L x 164 g/mol x 1L = 3.17 g

My answer was mix 3.17 g anhydrous sodium phosphate with 0.00207 L Phosphoric acid and Q.S. with water to 1 liter.

However, I was told by my instructor that this is incorrect, that the moles of acidic protons must be calculated. And while I have tried to figure this out, I keep getting stuck.

So the way I understand it is, Phosphoric Acid dissociates this way:

at low pH, phosphoric acid is completely protonated

at pH 2.15, there is 50% H3PO4 and 50% H2PO4 + [H+]

**at pH 7.2, there is 50% H2PO4- + [H+] and 50% HPO4 + 2[H+]**

at pH 12.35 there is 50% HPO4- + 2[H+] and 50% PO4- + 3[H+]

and at higher pH it is completely deprotonated.

Since the pka is at 7.2, and we want a pH of 7, for now let's say at pH 7 there is half the concentration at H2PO4 + [H+] and half HPO4- + 2[H+] (though I will want to know how to find the exact ratio of these two forms at pH 7 for the future, if there is a way to find this out, but for now, let's keep it simple)

Anhydrous Sodium phosphate has no acidic protons, and is in the form Na3PO4.

So for the buffer solution, there must be ratio [A-]/[HA] = 0.6310

now this is where my mind is hitting a wall:

I need to find the mols of acidic protons in order to create the ratio I calculated above, and 50% is H2PO4 + [H+] and 50% is HPO4 + 2[H+] and the second reagent has no acidic protons

.

my attempt:

14.8 mol/L H3PO4 x 3mol H / 1mol H3PO4 = 44.4M Hydrogen in 14.8 M H3PO4

so assuming there is 50:50 of the H2PO4 + [H+] and HPO4 2[H+]

14.8M [H+] in the first form, 29.6[H+] in the second,

0.5 (14.8M) + 0.5(29.6M) = 22.2 M acidic hydrogens at this pH

if [A-]/[HA]= 0.6310M

[A-]/[22.2]=0.6310M

[A-]= 14.008M

14.008mol/L Na3PO4 x 164 g/mol x1L = 2297.312 grams anhydrous sodium phosphate

if you need 22.2M acidic H in the phosphoric acid, and the ratio is 3molH /1mol H3PO4,

22.2mol/L H x 1molH3PO4/ 3mol H = 7.4 mol/L H3PO4

(14.8M)(x)=(7.4M) (1L)

0.578 L phosphoric acid

so add 2297.312 grams anhydrous sodium phosphate to 0.578 L phosphoric acid and Q.S. to 1L ?

But, if not all the Hydrogens are dissociated, is this the correct way to find the concentration of phosphoric acid?

for part (b), add 1mL HBr 8.89M, pka -8 and calculate the final pH

1mL x 1L/1000mL x 8.89 mol/L = 0.00889 mol HBr

so the ratio of the buffer solution is [A-]/[HA]= 0.6310M with 14.008 M [A-] anhydrous sodium phosphate, and 22.2M acidic H in H3PO4

subtract the mols of acid added from the amount of base,

14.008 mol/L x 1L = 14.008 mol [A-]

14.008 mol - 0.00889 mol HBr = 13.99911 mol

add the mols of acid added to the amount of acid,

22.2 mol/L x 1L = 22.2 mol [HA]

22.2 mol + 0.00889 mol HBr = 22.20889 mol

Using the Henderson Hasselbalch equation again, do I use the pka of the buffer solution?

so to find the pka of the buffer solution

pH=pka +log[A-]/[HA]

7=pka + log( .0.631)

7=pka - 0.19997

pka= 6.8

so

pH=pka+log[A-]/[HA]

pH=6.8 + log (13.99911/22.20889)

pH= 6.8 - 0.2004

pH= 6.6 after 1mL HBr added

I am pretty sure I calculated the quantity of acid wrong, somehow I am getting confused at comparing the [HA] from the H-H equation to the amount of [H+] dissociated between the two forms, any help would be much appreciated. Thank you!