Question

Describe the preparation of 100 ml of phosphate buffer pH 7.2 using a 0.5 M H2PO4 and O .5 M NaOH

Answer 1

I don't know what molarity you want the buffer to be. I have assume 1 M but that may not be suitable. I work in millimoles and not concentrations; however, since M = millimols/mL then the volume cancels (since it's the same solution) and that works out OK.
M buffer = 1 M or millimoles = mL x M = 100 x 1 = 100 mmols.
pH = pKa2 + log base/acid
7.2 = 7.2 + log base/acid
base/acid = 1 or b = a and that's equation 1.
b + a = 100 mmols = equation 2.
Solve these two equations simultaneously to get
b = 50 mmols and a = 50 mmols.
So you want to take some amount of 0.5 M H2PO4^- (say x mL), add some quantity of 0.5 M NaOH(say y mL) to convert it to HPO4^- and have the final volume = 100 mL(or x mL + y mL = 100 mL)
Start with x volume of H2PO4^- initially and add y of NaOH
...........H2PO4^- + OH^- ==> HPO4^2- + H2O
I.........0.5x..............0..................0....................
add.......................0.5y............................................
C......-0.5y...........-0.5y..............+0.5y
E........0.5x-0.5y........0...............0.5y

pH = pKa2 + log (b/a)
7.2 = 7.2 + log (0.5y/0.5x-0.5y)
1 = 0.5y/(0.5x-0.5y) or
y = x/2 but you want x mL + y mL = 100 so
Solve the two equations to get x = 66.67 mL H2PO4^- then you add 66.67/2 = ? mL NaOH. Check it.

...........H2PO4^- + OH^- ==> HPO4^2- + H2O
I.........0.5(66.67).......0..................0....................
............33.34 mmols
add.......................0.5(33.34).= 16.67..........................................
C......-16.67.............-16.67..............16.67
E.........16.67..............0.....................16.67
So base = acid which we calculated and total volume is 66.67 + 33.34 = 100 mL