Sodium Phosphate buffer has 3 ionizable groups, with pKa\'s of 2.15, 7.20 and 12

Question

Sodium Phosphate buffer has 3 ionizable groups, with pKa's of 2.15, 7.20 and 12.45. A 1.00 M solution of this buffer (200 mL) is found to have a pH of 7.50 A biochemist adds 30.0 mL of 1.00 M HCL to this buffer.

(a) How many moles of Na2HPO4 and NaH2PO4 are present before the addition of HCl?

(b) What is the resultant pH after the addition of HCL?

(C) How many total milliliters of 0.350 M NaOH can you add to the final solution in Question 8b and still maintain as a good phosphate buffer (i.e. within 1 pH unit of pKa).

Explanation / Answer

a) pH = pka + log [Na2HPO4]/[NaH2PO4]

7.5 = 7.2 + log [Na2HPO4]/[NaH2PO4]

[Na2HPO4] = 1.995 [NaH2PO4]

given buffer conc = 1 =[Na2HPO4]+[NaH2PO4]

solving above two we get [NaH2PO4 ] = 0.334

[Na2HPO4] = 0.666 , since vol = 200 ml = 0.2 liters

Na2HPO4 moles = 0.666 x0.2 = 0.133 ,

NaH2PO4 moles = 0.334 x0.2 = 0.0668,

b) H+ moles added = 0.03 x1 = 0.03 , Na2HPO4 + H+ <--------> NaH2PO4

now [Na2HPO4] = ( 0.133-0.03)/(0.2+0.03) = 0.448

[NaH2PO4] = ( 0.0688+0.03)/0.23 = 0.43

pH = 7.2 + log ( 0.448/0.43) = 7.22

C) pH of 1 unit can be tolerated , and hence pH limit of 7.2+1 = 8.2

now 8.2 = 7.22 + log [Na2HPO4]/[NaH2PO4]

[Na2HPO4] = 9.55 [NaH2PO4] , NaH2PO4 + NaOH <------> Na2HPO4 + H2O

let V = vol of NaOH

then ( 0.448 x 0.23 + 0.35V) = 9.55 ( 0.43 x 0.23 -0.35V)

0.103+ 0.35V = 0.9445-3.3425V

V = 0.228 liters = 228 ml of NaOH can be added

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